**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
#include "bits/stdc++.h"
using namespace std;
using ll = long long;

void solve()
{
	int n;
	cin >> n;
	vector<int> a(n);
	int mx = 0;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		mx = max(mx, a[i]);
	}
	vector<int> c(mx + 1);
	for (int i = 0; i < n; i++) {
		c[a[i]]++;
	}
	map<int, vector<int>> f;
	f[0].push_back(0);
	for(int i = 1; i <= mx; i++) {
		if (c[i]) {
			map<int, vector<int>> g = f;
			for (auto &[x, y]: f) {
				int gg = __gcd(x, i);
				if (!g.count(gg)) {
					auto p = y;
					p.push_back(i);
					g[gg] = p;
				}
			}
			f.swap(g);
		}
	}
	ll ans = 1e18;
	auto calc = [&](vector<int> y) {
		ll ret = 0;
		int gg = 0;
		auto b = c;
		for (auto &x : y) {
			gg = __gcd(x, gg);
			ret += gg;
			b[x]--;
		}
		for (int i = 1; i <= mx; i++) {
			if (b[i]) {
				gg = __gcd(i, gg);
				ret += 1ll * gg * b[i];
			}
		}
		return ret;
	};
	for (auto &[x, y]: f) {
		ans = min(ans, calc(y));
	}
	cout << ans << '\n';
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr), cout.tie(nullptr);
	int _ = 1;
	cin >> _;
	while (_--) {
		solve();
	}
	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the provided code, we need to consider edge cases and scenarios that can challenge the algorithm's assumptions and logic. Here are some key points to consider:

1. **Single Element Array**: The code should handle single-element arrays correctly.
2. **Arrays with Repeated Elements**: The code should correctly handle arrays where all elements are the same.
3. **Arrays with Coprime Elements**: The code should handle arrays where all elements are coprime (i.e., their GCD is 1).
4. **Arrays with Large Elements**: The code should handle large elements within the allowed range.
5. **Arrays with Mixed Elements**: The code should handle arrays with a mix of small and large elements.

Given these considerations, let's construct a small-scale test input that can expose potential bugs in the code.

### Test Input

```plaintext
3
1
10
3
4 4 4
5
6 7 8 9 10
```

### Explanation

1. **Single Element Array**:
   - Test case: `1 10`
   - Expected output: `10`
   - This tests the code's ability to handle a single-element array correctly.

2. **Arrays with Repeated Elements**:
   - Test case: `3 4 4 4`
   - Expected output: `12`
   - This tests the code's ability to handle arrays where all elements are the same.

3. **Arrays with Coprime Elements**:
   - Test case: `5 6 7 8 9 10`
   - Expected output: `5`
   - This tests the code's ability to handle arrays where all elements are coprime.

These test cases cover a range of scenarios that can help identify potential issues in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
/*   AUTHOR: Estufa en Piloto   */
#include <bits/stdc++.h>
using namespace std;

#ifdef LOCAL
#define DBG(x) cerr << #x << " = " << (x) << endl
#define RAYA cerr << "===============================" << endl
#else
#define DBG(x)
#define RAYA
#endif

typedef long long ll;
typedef vector<ll> vi; typedef pair<ll,ll> ii;
typedef vector<ii> vii; typedef vector<bool> vb;
#define FIN ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define forr(i, a, b) for(int i = (a); i < (ll) (b); i++)
#define forn(i, n) forr(i, 0, n)
#define SZ(x) int((x).size())
#define pb push_back
#define all(c) (c).begin(),(c).end()
#define esta(x,c) ((c).find(x) != (c).end())
const int INF = 1<<30; // const ll INF = (1LL<<60);
const int MOD = 1e9+7;  // const int MOD = 998244353;
const int MAXN  = 2e5+5;

// Criba lineal, obtiene los primos menores al parametro
vi min_prime; // min_prime[i] contiene el menor primo que divide a i, util para factorizar en log(i)

vi criba(ll n) {
    vb prime(n+1,true);
    min_prime.resize(n+1,INF);
    vi primos;
    for(ll p=2; p*p<=n; p++){
        if(!prime[p]) continue;
        for(ll i=p*p; i<=n; i += p) {
            prime[i] = false;
            min_prime[i] = min(min_prime[i],p);
        }
    }
    forr(i, 2, n+1){
        if(prime[i]) primos.pb(i), min_prime[i] = i;
    }
    return primos; // lista de primos hasta n
}

vi find_divisors(ll n) {
	vector <pair<ll,ll>> factor;
	while(n!=1) {
		int cont = 0; ll p=min_prime[n];
		while(n % p == 0) {
			cont++;
			n /= p;
		}
		factor.pb({p,cont});
	}

	vi divisores = {1};
	for(auto [p,exp] : factor) {
		int tam = SZ(divisores);
		forn(i,exp) {
			forn(j,tam) {
				int pos = SZ(divisores)-tam;
				divisores.pb(divisores[pos] * p);
			}
		}
	}

	sort(all(divisores));
	return divisores;
}

ll gcd(ll a, ll b){
	if(b==0) return a;
	else return gcd(b, a%b);
}

void solve(){
	int n; ll M=0, k;
	cin >> n;
	vi v(n);
	forn(i,n){
		cin >> v[i];
		M=max(M,v[i]);
		if(i==0) k=v[0];
		else k=gcd(k,v[i]);
	}
	forn(i,n) v[i]/=k;
	M/=k; M++;
	vi cont(M);
	vector <vi> d(n);
	forn(i,n){
		d[i]=find_divisors(v[i]);
		for(auto u : d[i]) cont[u]++;
	}
	vb haycop(M);
	forr(i,1,M){
		ll aux=i;
		vi primos;
		while(aux!=1) {
			ll p=min_prime[aux];
			while(aux % p == 0) aux /= p;
			primos.pb(p);
		}
		ll tot=0;
		forn(mask,1<<SZ(primos)){
			ll c=1, cnt=0;
			forn(j,SZ(primos)) if((mask&(1<<j))!=0){c*=primos[j]; cnt++;}
			if(cnt%2==0) tot-=cont[c];
			else tot+=cont[c];
		}
		if(tot!=0) haycop[i]=true;
	}
	ll ans=INF;
	forn(i,n){
		ll aux=n;
		for(int j=SZ(d[i])-1; j>=0; j--){
			if(v[i]%d[i][j]==0 and haycop[d[i][j]]){
				aux+=v[i]-1;
				v[i]/=d[i][j];
			}
		}
		ans=min(ans,aux);
	}
	cout << ans*k << "\n";
}

int main() {
	FIN;
	int t;
	cin >> t;
	criba(1e5);
	while(t--) solve();
	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
5
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
#include <bits/stdc++.h>

using namespace std;
#define ll long long int
#define all(a) (a).begin(),(a).end()
#define endl '\n'
#define d_g ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#ifndef ONLINE_JUDGE
#include "debug.hpp"
#else
#define debug(...) 69
#endif

        ll gcd(ll a,ll b)
        {
           return b==0 ? a : gcd(b,a%b);
        }

void solve()
{
    ll n;
    cin>>n;
    vector<ll> v(n+1ll,0ll);
    ll sum=0ll;
    for(ll i=1ll;i<=n;i++) cin>>v[i],sum=gcd(sum,v[i]);
    multiset<ll> s;
    for(ll i=1ll;i<=n;i++) v[i]/=sum,s.insert(v[i]);

    ll ans=*s.begin();
    s.erase(s.find(ans));
    ll best=ans;
    while(best>1)
    {
        ll as=-1ll;
        for(auto i:s)
        {
            ll x=gcd(i,ans);
            if(x<best)
            {
                best=x;
                as=i;
            }
        }

        if(as!=-1) s.erase(s.find(as)),   ans+=best;
    }
    ans+=s.size();
    ans*=sum;
    cout<<ans<<endl;

}

int main()
{
    d_g
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
    ll t=1ll;
    cin>>t;
    while(t--)
    {
        solve();
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
4
1 2 3 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
import math
t = int(input())
for _ in range(t):
    n = int(input())
    c = list(map(int, input().split()))
    c.sort()
    ngcd = c[0]
    ans = ngcd
    c.pop(0)
    r = 1
    while len(c) > 0 and ngcd > 1:
        mi = ngcd
        mn = ngcd
        ind = -1
        for i in range(0, len(c)):
            if math.gcd(ngcd, c[i]) <= mi:
                ind = i
            mn = math.gcd(mn, c[i])
            mi = min(mi, math.gcd(ngcd, c[i]))
        for i in range(0, len(c)):
            c[i] //= mn
        c.pop(ind)
        ngcd = mi
        ans += ngcd * r
        r *= mn
    ans += len(c) * r
    print(ans)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases that can cause the algorithm to fail. The main issue in the code is how it handles the GCD calculations and the way it modifies the array elements. Here are the key points to consider:

1. **Handling of GCD Calculation**: The code assumes that the GCD of the current element and the next element in the sorted list will always be minimized in a specific way. This assumption can be incorrect in certain cases, especially when the array contains many elements with a GCD of 1.

2. **Array Modification**: The code modifies the array elements by dividing them by the GCD, which can lead to incorrect results if the GCD is not properly handled.

3. **Edge Cases**: Consider arrays with all elements being the same, arrays with elements that are prime numbers, and arrays with a mix of elements that have different GCD properties.

Given these considerations, a small-scale test input that can expose the bug is:

```plaintext
3
5
1 2 3 4 5
3
1 1 1
4
2 4 6 8
```

### Explanation:
1. **Test Case 1:**
   - Array: `[1, 2, 3, 4, 5]`
   - The GCD of the entire array is 1. The expected output should be 15 (since each GCD of a single element is 1, and there are 5 elements).

2. **Test Case 2:**
   - Array: `[1, 1, 1]`
   - The GCD of the entire array is 1. The expected output should be 3 (since each GCD of a single element is 1, and there are 3 elements).

3. **Test Case 3:**
   - Array: `[2, 4, 6, 8]`
   - The GCD of the entire array is 2. The expected output should be 14 (since the GCDs are 2, 2, 2, and 2, and the sum is 2 + 2 + 2 + 2 = 8).

These test cases will help identify issues in the GCD calculation and array modification steps in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
#!/usr/bin/env python
import os
import sys
from math import gcd
from functools import reduce
from collections import Counter, defaultdict
from io import BytesIO, IOBase

def memodict(f):
    """memoization decorator for a function taking a single argument"""
    class memodict(dict):
        def __missing__(self, key):
            ret = self[key] = f(key)
            return ret

    return memodict().__getitem__

def pollard_rho(n):
    """returns a random factor of n"""
    if n & 1 == 0:
        return 2
    if n % 3 == 0:
        return 3

    s = ((n - 1) & (1 - n)).bit_length() - 1
    d = n >> s
    for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
        p = pow(a, d, n)
        if p == 1 or p == n - 1 or a % n == 0:
            continue
        for _ in range(s):
            prev = p
            p = (p * p) % n
            if p == 1:
                return gcd(prev - 1, n)
            if p == n - 1:
                break
        else:
            for i in range(2, n):
                x, y = i, (i * i + 1) % n
                f = gcd(abs(x - y), n)
                while f == 1:
                    x, y = (x * x + 1) % n, (y * y + 1) % n
                    y = (y * y + 1) % n
                    f = gcd(abs(x - y), n)
                if f != n:
                    return f
    return n

@memodict
def prime_factors(n):
    """returns a Counter of the prime factorization of n"""
    if n <= 1:
        return Counter()
    f = pollard_rho(n)
    return Counter([n]) if f == n else prime_factors(f) + prime_factors(n // f)

def distinct_factors(n):
    """returns a list of all distinct factors of n"""
    factors = [1]
    for p, exp in prime_factors(n).items():
        factors += [p**i * factor for factor in factors for i in range(1, exp + 1)]
    return factors

def main():
    t = int(input())

    for _ in range(t):
        n = int(input())
        a = {int(ai) for ai in input().split()}

        for i in list(a):
            for factor in distinct_factors(i):
                if factor != i and factor in a:
                    a.remove(i)
                    break

        cost = defaultdict(lambda: 10 ** 9)
        for ai in a:
            cost[ai] = ai

        for h in range(min(14, n - 1)):
            new_cost = defaultdict(lambda: 10 ** 9)
            for i, j in cost.items():
                for k in a:
                    curr = gcd(i, k)
                    new_cost[curr] = min(new_cost[curr], j + curr)
            cost = new_cost

            if h == n - 2:
                print(cost[min(cost)])
                break
        else:
            final = min(cost)
            base = cost[final]
            extra = n - 15
            print(base + final * extra)

# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion

if __name__ == "__main__":
    main()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, let's consider the following aspects:

1. **Array Size and Values**: The array size should be within the constraints (1 ≤ n ≤ 10^5) and the values should be within (1 ≤ a_i ≤ 10^5). We should focus on edge cases and scenarios that maximize the failure probability in faulty solutions.

2. **GCD Properties**: The GCD of a sequence can be significantly reduced by rearranging the elements. We should create an input where the GCD of the entire sequence is 1, but the intermediate GCDs are not trivial.

3. **Prime Numbers and Composite Numbers**: Using prime numbers and their multiples can help in creating a scenario where the GCD of the entire sequence is 1, but the intermediate GCDs are higher.

4. **Special Cases**: Consider arrays with repeated elements, arrays with a mix of prime and composite numbers, and arrays where the GCD of the entire sequence is 1.

Given these considerations, a challenging test input could be:

```plaintext
1
5
6 10 15 21 30
```

### Explanation:
- **Array**: [6, 10, 15, 21, 30]
- **GCD of the entire array**: gcd(6, 10, 15, 21, 30) = 1
- **Intermediate GCDs**:
  - gcd(6) = 6
  - gcd(6, 10) = 2
  - gcd(6, 10, 15) = 1
  - gcd(6, 10, 15, 21) = 1
  - gcd(6, 10, 15, 21, 30) = 1

This input is designed to test the algorithm's ability to handle a sequence where the GCD of the entire sequence is 1, but the intermediate GCDs are not trivial. The goal is to ensure that the algorithm correctly calculates the minimum possible value of the expression.

### Expected Output:
```plaintext
9
```

### Reasoning:
- The optimal rearrangement could be [6, 10, 15, 21, 30].
- The sum of the GCDs would be:
  - gcd(6) = 6
  - gcd(6, 10) = 2
  - gcd(6, 10, 15) = 1
  - gcd(6, 10, 15, 21) = 1
  - gcd(6, 10, 15, 21, 30) = 1
- Total: 6 + 2 + 1 + 1 + 1 = 11

However, the correct minimum value is 9, which can be achieved by a different rearrangement, such as [6, 15, 10, 21, 30]:
- gcd(6) = 6
- gcd(6, 15) = 3
- gcd(6, 15, 10) = 1
- gcd(6, 15, 10, 21) = 1
- gcd(6, 15, 10, 21, 30) = 1
- Total: 6 + 3 + 1 + 1 + 1 = 12

This input should expose any issues in the algorithm that fail to correctly minimize the sum of the GCDs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
import math

for _ in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))
    a.sort()
    ans = 0
    sc = 0
    gl = 1
    while True:
        g = a[0]
        s = set()
        for i in range(1, len(a)):
            g = math.gcd(g, a[i])
            s.add(math.gcd(a[0], a[i]))
        ans+=a[0]*gl
        gl*=g
        a = list(s)
        a.sort()
        for i in range(len(a)):
            a[i] = a[i]//g
        sc+=1
        if a[0]==1:
            ans+=gl*(n-sc)
            break
    print(ans)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
1
1
2
2 2
3
2 4 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
import java.io.*;
import java.util.*;

public class PrefixGCD {
    public static void main(String[] args) throws IOException {
        // Use BufferedReader for fast input
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        // Use BufferedWriter for fast output
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        StringTokenizer st;

        // Read number of test cases
        int t = Integer.parseInt(br.readLine());

        while (t-- > 0) {
            // Read n
            int n = Integer.parseInt(br.readLine());
            // Read array a[]
            long[] a = new long[n];
            st = new StringTokenizer(br.readLine());
            for (int i = 0; i < n; i++) {
                a[i] = Long.parseLong(st.nextToken());
            }

            // Compute GCD of all elements
            long G = a[0];
            for (int i = 1; i < n; i++) {
                G = gcd(G, a[i]);
                if (G == 1)
                    break; // Early termination if GCD becomes 1
            }

            // Compute array b[] = a[i] / G
            long[] b = new long[n];
            for (int i = 0; i < n; i++) {
                b[i] = a[i] / G;
            }

            // Find the minimal bi_min
            int bi_min_index = 0;
            for (int i = 1; i < n; i++) {
                if (b[i] < b[bi_min_index]) {
                    bi_min_index = i;
                }
            }
            long bi_min = b[bi_min_index];
            // Initialize sum
            long sum = bi_min * G;
            // Prepare the list of remaining b's
            ArrayList<Long> sorted_b = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                if (i != bi_min_index) {
                    sorted_b.add(b[i]);
                }
            }
            // Sort the remaining b's in ascending order
            Collections.sort(sorted_b);

            // Initialize current GCD
            long currentGCD = bi_min;

            // Initialize pointer
            int index = 0;

            while (index < sorted_b.size()) {
                if (currentGCD == 1) {
                    // Add G for all remaining elements
                    sum += G * (sorted_b.size() - index);
                    break;
                } else {
                    // Try to find an element with GCD(currentGCD, bi) ==1
                    int found = -1;
                    for (int j = index; j < sorted_b.size(); j++) {
                        if (gcd(currentGCD, sorted_b.get(j)) == 1) {
                            found = j;
                            break;
                        }
                    }
                    if (found != -1) {
                        // Add G to sum
                        sum += G;
                        // Update current GCD to 1
                        currentGCD = 1;
                        // Remove the found element by swapping with current index
                        Collections.swap(sorted_b, found, index);
                        // Increment index as the element at 'index' is now used
                        index++;
                    } else {
                        // Choose the first bi, update current GCD
                        long bi = sorted_b.get(index);
                        long ci = gcd(currentGCD, bi);
                        sum += ci * G;
                        currentGCD = ci;
                        index++;
                    }
                }
            }

            // Append the sum to the output
            bw.write(String.valueOf(sum));
            bw.newLine();
        }

        // Flush and close the output
        bw.flush();
        bw.close();
        br.close();
    }

    // Method to compute GCD using Euclidean algorithm
    private static long gcd(long a, long b) {
        while (b != 0) {
            long temp = a % b;
            a = b;
            b = temp;
        }
        return a;
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are the steps to consider:

1. **Edge Case: All Elements are the Same**:
   - If all elements in the array are the same, the GCD of the entire array is the same element. This can cause issues if the code does not handle this case correctly.

2. **Edge Case: Elements with a Common Factor**:
   - If all elements have a common factor greater than 1, the GCD of the entire array will be that common factor. This can expose issues in how the code handles common factors.

3. **Edge Case: Elements with a GCD of 1**:
   - If the GCD of the entire array is 1, the code should handle this efficiently. This can test the early termination and the logic for handling GCD of 1.

4. **Edge Case: Large Array Size**:
   - While the input size is limited, a larger array can still stress-test the algorithm and expose inefficiencies.

5. **Edge Case: Mixed Values**:
   - An array with mixed values, including both large and small numbers, can test the robustness of the GCD calculation and the sorting logic.

Given these considerations, let's create a small-scale test input that includes these edge cases:

```plaintext
5
3
4 4 4
3
4 6 8
3
7 11 13
5
1 2 3 4 5
4
2 4 6 8
```

### Explanation of the Test Input:
1. **Test Case 1**: All elements are the same (4, 4, 4).
   - Expected Output: 12 (4 + 4 + 4)

2. **Test Case 2**: Elements with a common factor (4, 6, 8).
   - Expected Output: 12 (4 + 2 + 2)

3. **Test Case 3**: Elements with a GCD of 1 (7, 11, 13).
   - Expected Output: 21 (7 + 1 + 1)

4. **Test Case 4**: A larger array with mixed values (1, 2, 3, 4, 5).
   - Expected Output: 15 (1 + 1 + 1 + 1 + 1)

5. **Test Case 5**: Elements with a common factor (2, 4, 6, 8).
   - Expected Output: 20 (2 + 2 + 2 + 2)

This test input should help expose any bugs in the provided code, especially in handling edge cases and common factors.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
import java.util.*;

public class Main {
    static final int N = 1001000;

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int T = scanner.nextInt();

        while (T-- > 0) {
            int n = scanner.nextInt();
            int[] a = new int[N];

            for (int i = 1; i <= n; i++) {
                a[i] = scanner.nextInt();
            }

            int G = 0;
            for (int i = 1; i <= n; i++) {
                G = gcd(a[i], G);
            }

            Arrays.sort(a, 1, n + 1);

            int ans = a[1];
            int g = a[1];
            int rm = n - 1;
            a[1] = 0;

            while (g != G) {
                int mn = (int)1e9;
                int p = 0;

                for (int i = 1; i <= n; i++) {
                    if (a[i] != 0 && gcd(a[i], g) < mn) {
                        mn = gcd(a[i], g);
                        p = i;
                    }
                }

                ans += mn;
                g = mn;
                rm--;
                a[p] = 0;
            }

            ans += g * rm;
            System.out.println(ans);
        }

        scanner.close();
    }

    static int gcd(int x, int y) {
        return y != 0 ? gcd(y, x % y) : x;
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the provided code, we need to consider edge cases or scenarios that might cause the algorithm to fail. The code is supposed to find the smallest possible value of the expression \(\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n)\).

Here are some potential edge cases to consider:
1. **All elements are the same**: This could expose issues with the GCD calculation.
2. **Elements are all prime numbers**: This could test the GCD calculation and the selection process.
3. **Elements are a mix of small and large numbers**: This could test the sorting and GCD calculation.
4. **Elements are in a specific order that might cause the algorithm to make suboptimal choices**.

Given these considerations, let's create a test input that includes a mix of small and large numbers, and also includes a specific order that might cause the algorithm to fail.

```plaintext
1
5
100 2 50 4 1
```

This test case includes:
- A large number (100) and a small number (1).
- A mix of other numbers (50, 4, 2).

The expected output for this test case should be 14, which is the minimum possible value for the given array. The correct arrangement could be [1, 2, 4, 50, 100], resulting in:
- \(\gcd(1) = 1\)
- \(\gcd(1, 2) = 1\)
- \(\gcd(1, 2, 4) = 1\)
- \(\gcd(1, 2, 4, 50) = 1\)
- \(\gcd(1, 2, 4, 50, 100) = 1\)

Summing these values gives \(1 + 1 + 1 + 1 + 10 = 14\).

This test case should expose any issues in the algorithm's handling of large and small numbers, and the order in which elements are processed.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        vector<ll> a(n);
        for(auto &x:a) cin>>x;
        if(n ==1){
            cout<<a[0]<<"\n";
            continue;
        }

        ll g = a[0];
        for(int i=1;i<n;i++) g = gcd(g, a[i]);

        ll a_min = *min_element(a.begin(), a.end());

        int pos = -1;
        for(int i=0;i<n;i++) if(a[i] == a_min){ pos = i; break;}

        vector<ll> b;
        for(int i=0;i<n;i++) if(i != pos) b.push_back(a[i]);

        sort(b.begin(), b.end(), [&](const ll &x, const ll &y) -> bool{
            ll gx = gcd(a_min, x);
            ll gy = gcd(a_min, y);
            return gx < gy;
        });

        vector<ll> arranged;
        arranged.push_back(a_min);
        for(auto &x:b) arranged.push_back(x);

        ll current_gcd = arranged[0];
        ll sum = current_gcd;
        for(int i=1;i<n;i++){
            current_gcd = gcd(current_gcd, arranged[i]);
            sum += current_gcd;
        }
        cout<<sum<<"\n";
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases that might cause the algorithm to fail. The code attempts to minimize the sum of GCDs by sorting the elements in a specific way. However, this approach might not always yield the correct result, especially in cases where the minimum element does not play a significant role in minimizing the GCDs.

Let's consider the following edge cases:
1. **Array with all elements being the same**: This is a simple case where the GCD of any subset of the array is the element itself.
2. **Array with a small minimum element and large other elements**: This can test how the code handles the sorting and GCD calculations.
3. **Array with a mix of small and large elements**: This can test the code's ability to handle a more complex arrangement.

Given the constraints, we will generate a small-scale test input that is complete and valid.

Here is a test input that can expose potential issues in the code:

```plaintext
3
3
2 2 2
5
1 10 100 1000 10000
4
1 2 3 4
```

Explanation:
1. **Test Case 1**: All elements are the same. The expected output is `6` because `gcd(2) + gcd(2, 2) + gcd(2, 2, 2) = 2 + 2 + 2 = 6`.
2. **Test Case 2**: The array has a small minimum element (1) and large other elements. The expected output is `1 + 1 + 1 + 1 + 1 = 5`.
3. **Test Case 3**: The array has a mix of small and large elements. The expected output is `1 + 1 + 1 + 1 = 4`.

This test input should help in identifying any issues with the sorting and GCD calculation logic in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;

using ll = long long;

void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    int min_a = numeric_limits<int>::max();
    for (int &x : a) {
        cin >> x;
        min_a = min(min_a, x);
    }
    int G = min_a;
    ll sum = G;
    vector<pair<int, int>> gcds;
    bool skipped_min_a = false;
    for (int x : a) {
        if (x == min_a && !skipped_min_a) {
            skipped_min_a = true;
            continue;
        }
        int g = gcd(G, x);
        gcds.emplace_back(g, x);
    }
    sort(gcds.begin(), gcds.end());
    for (auto &p : gcds) {
        int x = p.second;
        G = gcd(G, x);
        sum += G;
    }
    cout << sum << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider the following points:

1. **Edge Cases**: The code might fail for edge cases where the array has only one element, all elements are the same, or the elements are very small.
2. **Behavior with Minimum Element**: The code skips the minimum element once, which might cause issues if the minimum element is the only element that can reduce the GCD significantly.
3. **Order of Elements**: The code sorts the elements based on the GCD, which might not always be the optimal order to minimize the sum of GCDs.

Let's construct a test input that can expose these issues:

1. **Single Element Array**: This should be handled trivially, but it's a good edge case to test.
2. **All Elements the Same**: This should also be handled trivially, but it's a good edge case to test.
3. **Minimum Element is the Only Element that Reduces GCD**: This will test if the code correctly handles the minimum element.
4. **Optimal Order of Elements**: This will test if the code correctly handles the order of elements to minimize the sum of GCDs.

Here is a small-scale test input that covers these points:

```plaintext
4
1
5
3
1 1 1
5
2 3 6 6 6
4
2 4 6 12
```

### Explanation:

1. **Test Case 1**: Single element array.
   - Input: `1 5`
   - Expected Output: `5` (since GCD(5) = 5)

2. **Test Case 2**: All elements are the same.
   - Input: `3 1 1 1`
   - Expected Output: `3` (since GCD(1) + GCD(1, 1) + GCD(1, 1, 1) = 1 + 1 + 1 = 3)

3. **Test Case 3**: Minimum element is the only element that reduces GCD.
   - Input: `5 2 3 6 6 6`
   - Expected Output: `10` (since GCD(2) + GCD(2, 3) + GCD(2, 3, 6) + GCD(2, 3, 6, 6) + GCD(2, 3, 6, 6, 6) = 2 + 1 + 1 + 1 + 1 = 10)

4. **Test Case 4**: Optimal order of elements.
   - Input: `4 2 4 6 12`
   - Expected Output: `14` (since GCD(2) + GCD(2, 4) + GCD(2, 4, 6) + GCD(2, 4, 6, 12) = 2 + 2 + 2 + 2 = 14)

This test input should help expose any issues with the provided code, especially in handling the minimum element and the order of elements.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
import math
t=int(input())
for i in range(t):
    n=int(input())
    a=list(map(int,input().split()))
    a.sort()
    if len(a)<2:
        print(a[0])
    else:
        m=math.gcd(a[0],a[1])
        ans=1
        for i in range(2,len(a)):
            # m=min(m,math.gcd(a[0],a[i]))
            b=math.gcd(a[0],a[i])
            if b<m:
                ans=i
        if ans!=1:
            a[1],a[ans]=a[ans],a[1]
        ans=0
        ans+=a[0]
        ans+=math.gcd(a[0],a[1])
        gcd=math.gcd(a[0],a[1])
        for i in range(2,len(a)):
            ans+=math.gcd(gcd,a[i])
            gcd=math.gcd(gcd,a[i])
        print(ans)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
5
1
1
2
1 2
3
1 2 3
4
1 2 3 4
5
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
# AI testing

import sys
import math
import threading

def main():
    def solve():
        import sys

        import sys

        t_and_rest = sys.stdin.read().split()
        ptr = 0
        if ptr >= len(t_and_rest):
            return
        t = int(t_and_rest[ptr])
        ptr +=1
        for _ in range(t):
            if ptr >= len(t_and_rest):
                break
            n = int(t_and_rest[ptr])
            ptr +=1
            a = []
            for _ in range(n):
                if ptr >= len(t_and_rest):
                    break
                a.append(int(t_and_rest[ptr]))
                ptr +=1
            if len(a)!=n:
                # Read remaining
                a += list(map(int, sys.stdin.readline().strip().split()))
            # Find the smallest element
            a_min = min(a)
            # Find all elements except a_min
            # If multiple a_min, remove one
            a_rest = []
            min_found = False
            for num in a:
                if num == a_min and not min_found:
                    min_found = True
                else:
                    a_rest.append(num)
            # Sort a_rest based on gcd(a_min, ai) in ascending order
            a_rest_sorted = sorted(a_rest, key=lambda x: math.gcd(a_min, x))
            # Now, compute the sum
            current_gcd = a_min
            total_sum = a_min
            for num in a_rest_sorted:
                current_gcd = math.gcd(current_gcd, num)
                total_sum += current_gcd
            print(total_sum)
    threading.Thread(target=solve).start()

if __name__ == "__main__":
    main()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
1
1
2
1 2
3
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        long t = scanner.nextLong();
        ArrayList<Long> arr;
        int n;
        while(t-- > 0){
            n = scanner.nextInt();
            arr = new ArrayList<Long>();
            long min = (long)Math.pow(10, 5);
            int ind = 0;
            while(ind < n){
                arr.add(scanner.nextLong());
                min = Math.min(min, arr.get(ind++));
            }
            long sum = min;
            long minGCD = min;
            ArrayList<Long> gcdArr = new ArrayList<Long>();

            while(ind-- > 0){
                long gcd = gcd(min, arr.get(ind));
                gcdArr.add(gcd);
            }
            gcdArr.sort(null);
            long currGcd = min;
            while(++ind < n-1){
                long temp = gcd(currGcd, gcdArr.get(ind));
                sum += temp;
                currGcd = temp;
            }
            System.out.println(sum);
        }
    }

    static long gcd(long a, long b)
    {
        if (b == 0)
            return a;

        else
            return gcd(b, a % b);
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that can maximize the failure probability. The primary issues in the code are:

1. The code incorrectly assumes that the minimum value in the array is always the starting point for the GCD calculation.
2. The code does not correctly handle the order of elements, which is crucial for the problem.

Given these issues, we can create a test case where the minimum value is not the best starting point for the GCD calculation. A good test case would be one where the minimum value is not the best choice for minimizing the sum of GCDs.

Let's consider the following test case:

```plaintext
1
4
2 4 6 12
```

### Explanation:
- **Input:**
  - Number of test cases: 1
  - Array size: 4
  - Array elements: 2, 4, 6, 12

### Expected Output:
- The minimum possible value of the expression is 10.

### Reasoning:
- If we start with the minimum value 2, the GCDs would be:
  - \(\gcd(2) = 2\)
  - \(\gcd(2, 4) = 2\)
  - \(\gcd(2, 4, 6) = 2\)
  - \(\gcd(2, 4, 6, 12) = 2\)
  - Sum = 2 + 2 + 2 + 2 = 8

- However, if we rearrange the array to [12, 6, 4, 2], the GCDs would be:
  - \(\gcd(12) = 12\)
  - \(\gcd(12, 6) = 6\)
  - \(\gcd(12, 6, 4) = 2\)
  - \(\gcd(12, 6, 4, 2) = 2\)
  - Sum = 12 + 6 + 2 + 2 = 22

- The correct minimum sum is actually 10, which can be achieved by rearranging the array to [6, 4, 2, 12]:
  - \(\gcd(6) = 6\)
  - \(\gcd(6, 4) = 2\)
  - \(\gcd(6, 4, 2) = 2\)
  - \(\gcd(6, 4, 2, 12) = 2\)
  - Sum = 6 + 2 + 2 + 2 = 12

### Test Input:
```plaintext
1
4
2 4 6 12
```

This test case will expose the bug in the code because the code incorrectly assumes that starting with the minimum value is always optimal, which is not the case here.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
import java.io.*;
import java.util.*;

public class Main {
    static int N = 200010;
    static int[] a = new int[N];

    static int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    public static void main(String[] args) throws IOException {
        FastScanner f = new FastScanner();
        PrintWriter w = new PrintWriter(System.out);
        int T = f.nextInt();
        while (T-- > 0) {
            int n = f.nextInt();
            //gcd每次都会让答案变小 2倍以上所以 其实gcd不为1的数的前缀不超过 logn 差不多 15就行
            //可以暴力去找让当前前缀最小的数拿出来就行
            for (int i = 1; i <= n; i++) a[i] = f.nextInt();
            int pre = 0x3f3f3f3f;
            long ans = 0;
            for (int i = 1; i <= Math.min(n, 15); i++) {
                int res = 0x3f3f3f3f;
                int id = -1;
                for (int j = 1; j <= n; j++) {
                    if (a[j] == -1) continue;
                    if (i == 1) {
                        if (res > a[j]) {
                            res = a[j];
                            id = j;
                        }
                    } else {
                        if (gcd(pre, a[j]) < res) {
                            res = gcd(pre, a[j]);
                            id = j;
                        }
                    }
                }
                if (id > 0) a[id] = -1;
                if (res == 1) {
                    ans += (n - i + 1);
                    break;
                } else ans += res;
                pre = res;
            }
            if (pre > 1) {
                long cnt = Math.max(0, n - 15);
                ans += cnt * pre;
            }
            w.println(ans);
        }
        w.flush();
    }

    private static class FastScanner {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        private FastScanner() throws IOException {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        private short nextShort() throws IOException {
            short ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do ret = (short) (ret * 10 + c - '0');
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return (short) -ret;
            return ret;
        }

        private int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do ret = ret * 10 + c - '0';
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return -ret;
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do ret = ret * 10 + c - '0';
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return -ret;
            return ret;
        }

        private char nextChar() throws IOException {
            byte c = read();
            while (c <= ' ') c = read();
            return (char) c;
        }

        private String nextString() throws IOException {
            StringBuilder ret = new StringBuilder();
            byte c = read();
            while (c <= ' ') c = read();
            do {
                ret.append((char) c);
            } while ((c = read()) > ' ');
            return ret.toString();
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1) buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead) fillBuffer();
            return buffer[bufferPointer++];
        }
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, let's consider the following steps:

1. **Understand the Bug**: The code tries to minimize the sum of GCDs by selecting the smallest possible GCDs for the first few elements. However, it assumes that once the GCD becomes 1, the remaining elements will not affect the result. This assumption can lead to incorrect results if the initial elements are not chosen optimally.

2. **Edge Cases**:
   - **Small Array with High GCDs**: If the array has high GCDs, the code might not handle it correctly.
   - **Array with Many 1s**: If the array contains many 1s, the code might not handle the remaining elements correctly.
   - **Array with Repeated Elements**: If the array has repeated elements, the code might not handle the GCDs correctly.

3. **Generate Test Input**:
   - **Test Case 1**: Small array with high GCDs.
   - **Test Case 2**: Array with many 1s.
   - **Test Case 3**: Array with repeated elements.

Let's create a test input that combines these edge cases:

```plaintext
3
3
10 10 10
5
1 1 1 1 1
4
6 6 6 6
```

### Explanation:
1. **Test Case 1**: Array `[10, 10, 10]`
   - The GCD of any prefix will be 10, and the sum of GCDs will be `10 + 10 + 10 = 30`.

2. **Test Case 2**: Array `[1, 1, 1, 1, 1]`
   - The GCD of any prefix will be 1, and the sum of GCDs will be `1 + 1 + 1 + 1 + 1 = 5`.

3. **Test Case 3**: Array `[6, 6, 6, 6]`
   - The GCD of any prefix will be 6, and the sum of GCDs will be `6 + 6 + 6 + 6 = 24`.

These test cases are designed to challenge the code's assumptions and edge cases, particularly when the GCDs are high or when the array contains many 1s or repeated elements.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
def gcd(a, b):
    while b:
        a, b = b, a % b
    return a

def main():
    n = int(input())  # Number of test cases

    for _ in range(n):
        x = int(input())  # Number of elements
        values = list(map(int, input().split()))  # Read all values in one line

        total_sum = 0
        prev = 0  # Initialize prev

        processed = [False] * x  # Track processed elements

        for _ in range(x):
            min_gcd = float('inf')
            min_index = -1

            # Find the unprocessed element with the minimum GCD
            for j in range(x):
                if not processed[j]:
                    current_gcd = gcd(prev, values[j])
                    if current_gcd < min_gcd:
                        min_gcd = current_gcd
                        min_index = j

            # Mark the selected element as processed
            processed[min_index] = True
            total_sum += min_gcd
            prev = min_gcd  # Update prev to the current min_value

        print(total_sum)

if __name__ == "__main__":
    main()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
5
3
1 2 3
2
1 1
3
2 4 6
4
1 2 3 4
5
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
from sys import stdin
input = lambda:stdin.readline().rstrip()
N = int(input())
def gcd(a,b):
        a,b = max(a,b), min(a,b)
        while b != 0:
            a,b = b, a % b
        return a
ans = [0]*N
for k in range(N):
    n = int(input())
    L = [int(i) for i in input().split()]
    lastGcd = 0
    totGcd = 0
    for _ in range(n):
        choice = min(L, key = lambda x : gcd(x,lastGcd))
        lastGcd = gcd(choice, lastGcd)
        totGcd += lastGcd
        L.remove(choice)
    ans[k] = totGcd
print("\n".join(str(i) for i in ans))

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
5
3
1 1 1
2
1 2
4
2 2 2 2
5
1 2 3 4 5
6
10 20 30 40 50 60
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
import java.util.*;

public class Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        while (T-- > 0) {
            int n = sc.nextInt();
            int[] a = new int[n];
            for (int j = 0; j < n; j++) {
                a[j] = sc.nextInt();
            }
            Arrays.sort(a);
            boolean[] used = new boolean[n];
            int cur = a[0];
            int sum = cur;
            used[0] = true;
            for (int i = 1; i < n; i++){
                int next = -1;
                int min = Integer.MAX_VALUE;
                for (int j = 0; j < n; j++) {
                    if (!used[j]) {
                        int temp = gcd(cur, a[j]);
                        if (temp < min) {
                            min = temp;
                            next = j;
                        }
                    }
                }
                used[next] = true;
                cur = min;
                sum += cur;
            }
            System.out.println(sum);
        }
    }

    public static int gcd(int a, int b) {
        if(a == 0) return b;
        return gcd(b%a, a);
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
5
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define vec vector<int>
#define vll vector<long long>
#define ll long long
#define endl "\n"

typedef unsigned long long ull;
typedef long double lld;

#ifndef ONLINE_JUDGE
#define debug(x)           \
    {                      \
        cerr << #x << ' '; \
        _print(x);         \
        cerr << endl;      \
    }
#else
#define debug(x)
#endif
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(lld t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(ull t) { cerr << t; }

mt19937_64 RNG(chrono::steady_clock::now().time_since_epoch().count());

const long long constraint = 1000000000000000000LL; // 10^18
const int N = 3e5 + 10;
const int mod = 1e9 + 7;

int solve()
{

    int n;
    cin >> n;

    multiset<int> st;
    set<int> sst;
    for (int i = 0; i < n; i++)
    {
        int x;
        cin >> x;
        st.insert(x);
    }
    vector<int> a;
    int sum = 0;
    int first = *st.begin();
    st.erase(st.begin());
    a.push_back(first);
    sum += first;
    for (auto ele : st)
    {
        sst.insert(gcd(first, ele));
    }
    int g = first;
    for (auto ele : sst)
    {
        g = gcd(g, ele);
        a.push_back(g);
        sum += g;
    }
    int count = a.size();
    int last = a[a.size() - 1];
    cout << sum + (n - count) * g << endl;

    return 0;
}

signed main()
{
    auto begin = std::chrono::high_resolution_clock::now();

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

    int t;
    cin >> t;
    // t = 1;

    while (t--)
    {
        // debug(t);

        solve();
    }
    auto end = std::chrono::high_resolution_clock::now();
    auto elapsed = std::chrono::duration_cast<std::chrono::nanoseconds>(end - begin);
    cerr << "Time measured: " << elapsed.count() * 1e-9 << " seconds.\n";
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases where the algorithm might fail. The buggy code has a few potential issues:

1. **Handling of GCD Calculation**: The code uses a set to store the GCDs of the first element with all other elements. This can lead to incorrect results if the GCDs are not calculated in the correct order.
2. **Handling of Duplicates**: The code uses a multiset to store the input elements, but it does not handle duplicates correctly in the GCD calculation.

Let's create a test case that can expose these issues. A good test case would be one where the input array has multiple elements with a common divisor, and the order of elements matters.

### Test Case
We will use an array where the elements have a common divisor, and the order of elements can affect the GCD calculation.

```plaintext
1
5
12 18 24 30 36
```

### Explanation
- **Input**: The array `[12, 18, 24, 30, 36]` has a common divisor of 6.
- **Expected Output**: The smallest possible value of the expression is 21.

Let's break down the expected output:
- \(\gcd(12) = 12\)
- \(\gcd(12, 18) = 6\)
- \(\gcd(12, 18, 24) = 6\)
- \(\gcd(12, 18, 24, 30) = 6\)
- \(\gcd(12, 18, 24, 30, 36) = 6\)

So, the sum is \(12 + 6 + 6 + 6 + 6 = 36\).

However, the correct approach should be to rearrange the array to minimize the sum. One possible rearrangement is `[6, 12, 18, 24, 30]`:
- \(\gcd(6) = 6\)
- \(\gcd(6, 12) = 6\)
- \(\gcd(6, 12, 18) = 6\)
- \(\gcd(6, 12, 18, 24) = 6\)
- \(\gcd(6, 12, 18, 24, 30) = 6\)

So, the sum is \(6 + 6 + 6 + 6 + 6 = 30\).

The buggy code might not handle this correctly, leading to an incorrect output.

### Output
```plaintext
1
5
12 18 24 30 36
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
#include <iostream>
#include <vector>
#include <random>
#include <algorithm>

using namespace std;

int gcd(int a, int b) {
	if (b == 0) {
		return a;
	}
	return gcd(b, a % b);
}

random_device rd;
mt19937 generator(rd());

int calculate(vector<int> &a) {
	int sum = 0;
	int g = 0;
	for (auto &i: a) {
		g = gcd(g, i);
		sum += g;
	}
	return sum;
}

void solve() {
	int n;
	cin >> n;
	vector <int> a(n);
	for (auto &i: a) {
		cin >> i;
	}
	int result = calculate(a);
	for (int i = 0; i < 1000; ++i) {
		shuffle(a.begin(), a.end(), generator);
		result = min(result, calculate(a));
	}
	sort(a.begin(), a.end());
	result = min(result, calculate(a));
	cout << result << "\n";
}

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cout.setf(ios::fixed);
	cout.precision(10);
	int n;
	cin >> n;
	while (n > 0) {
		--n;
		solve();
	}
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to consider the following:

1. **Edge Cases**: The code should handle edge cases where the array has very specific properties that might not be well-handled by the random shuffling and sorting approach.
2. **Special Values**: Values that are prime or have specific GCD properties.
3. **Multiple Test Cases**: To ensure the code is tested under various conditions.

Let's think about a test case that could expose the bug:

- **Array with All Elements the Same**: This should be easy to handle, but it's a good starting point.
- **Array with Elements That Have a Common Factor**: This could expose issues with the GCD calculation.
- **Array with Elements That Have No Common Factor**: This could expose issues with the random shuffling approach.
- **Array with a Mix of Large and Small Values**: This could expose issues with the sorting and shuffling approach.

Let's create a test input that includes these scenarios:

```plaintext
4
3
4 4 4
4
6 15 10 2
5
10 20 30 40 50
6
2 3 5 7 11 13
```

### Explanation:
1. **First Test Case (All Elements the Same)**:
   - Array: [4, 4, 4]
   - Expected Output: 12 (since GCD(4) + GCD(4, 4) + GCD(4, 4, 4) = 4 + 4 + 4 = 12)

2. **Second Test Case (Common Factor)**:
   - Array: [6, 15, 10, 2]
   - Expected Output: 9 (since one possible arrangement is [2, 6, 10, 15], and GCD(2) + GCD(2, 6) + GCD(2, 6, 10) + GCD(2, 6, 10, 15) = 2 + 2 + 2 + 1 = 9)

3. **Third Test Case (Multiple of 10)**:
   - Array: [10, 20, 30, 40, 50]
   - Expected Output: 15 (since one possible arrangement is [10, 20, 30, 40, 50], and GCD(10) + GCD(10, 20) + GCD(10, 20, 30) + GCD(10, 20, 30, 40) + GCD(10, 20, 30, 40, 50) = 10 + 10 + 10 + 10 + 10 = 15)

4. **Fourth Test Case (No Common Factor)**:
   - Array: [2, 3, 5, 7, 11, 13]
   - Expected Output: 6 (since GCD(2) + GCD(2, 3) + GCD(2, 3, 5) + GCD(2, 3, 5, 7) + GCD(2, 3, 5, 7, 11) + GCD(2, 3, 5, 7, 11, 13) = 2 + 1 + 1 + 1 + 1 + 1 = 6)

This test input should help expose the bug in the provided buggy code by covering a range of scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Prefix GCD
time_limit_ms: 2000
memory_limit_mb: 256
Description: Since Mansur is tired of making legends, there will be no legends for this task.

You are given an array of positive integer numbers $a_1, a_2, \ldots, a_n$. The elements of the array can be rearranged in any order. You need to find the smallest possible value of the expression $$\gcd(a_1) + \gcd(a_1, a_2) + \ldots + \gcd(a_1, a_2, \ldots, a_n),$$ where $\gcd(a_1, a_2, \ldots, a_n)$ denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $a_1, a_2, \ldots, a_n$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single number $n$ ($1 \le n \le 10^5$) — the size of the array.

The second line of each test case contains $n$ numbers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^5$) — the initial array.

The sum of $n$ over all test cases does not exceed $10^5$.

The sum of $\max(a_1, a_2, \ldots, a_n)$ over all test cases does not exceed $10^5$.

Output: For each test case, output a single number on a separate line — the answer to the problem.

Examples:
input:
5
3
4 2 2
2
6 3
3
10 15 6
5
6 42 12 52 20
4
42 154 231 66
output:
6
6
9
14
51

Note: In the first test case, the elements can be rearranged as follows: $[2, 4, 2]$. Then the answer will be $\gcd(2) + \gcd(2, 4) + \gcd(2, 4, 2) = 2 + 2 + 2 = 6$.

In the third test case, the elements can be rearranged as follows: $[6, 10, 15]$. Then the answer will be $\gcd(6) + \gcd(6, 10) + \gcd(6, 10, 15) = 6 + 2 + 1 = 9$.

Buggy Code:
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.util.Locale;
import java.util.StringTokenizer;

public class Solution implements Runnable {

	private PrintStream out;
	private BufferedReader in;
	private StringTokenizer st;

	public void solve() throws IOException {
		long time0 = System.currentTimeMillis();

		int t = nextInt();
		for (int test = 1; test <= t; test++) {
			int n = nextInt();
			int[] a = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = nextInt();
			}
			long answer = solve(n, a);
			out.println(answer);
		}

		System.err.println("time: " + (System.currentTimeMillis() - time0));
	}

	private long solve(int n, int[] a) {
		long answer = 0;
		for (int i = 0; i < n; i++) {
			for (int j = i + 1; j < n; j++) {
				if (a[i] > a[j]) {
					int tmp = a[i];
					a[i] = a[j];
					a[j] = tmp;
				}
			}
			answer = answer + a[i];
			boolean hasno = false;
			for (int j = i + 1; j < n; j++) {
				a[j] = gcd(a[j], a[i]);
				if (a[j] > 1) {
					hasno = true;
				}
			}
			if (!hasno) {
				answer = answer + n - 1 - i;
				break;
			}
		}
		return answer;
	}

	private int gcd(int i, int j) {
		while (i > 0) {
			int tmp = i;
			i = j % i;
			j = tmp;
		}
		return j;
	}

	public double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}

	public long nextLong() throws IOException {
		return Long.parseLong(next());
	}

	public int nextInt() throws IOException {
		return Integer.parseInt(next());
	}

	public String next() throws IOException {
		while (!st.hasMoreTokens()) {
			String line = in.readLine();
			if (line == null) {
				return null;
			}
			st = new StringTokenizer(line);
		}
		return st.nextToken();
	}

	@Override
	public void run() {
		try {
			solve();
			out.close();
		} catch (Throwable e) {
			throw new RuntimeException(e);
		}
	}

	public Solution(String name) throws IOException {
		Locale.setDefault(Locale.US);
		if (name == null) {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintStream(new BufferedOutputStream(System.out));
		} else {
			in = new BufferedReader(new InputStreamReader(new FileInputStream(name + ".in")));
			out = new PrintStream(new BufferedOutputStream(new FileOutputStream(name + ".out")));
		}
		st = new StringTokenizer("");
	}

	public static void main(String[] args) throws IOException {
		new Thread(new Solution(null)).start();
	}
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the given code, we need to consider the logic and how it might fail. The code attempts to sort the array and then compute the GCD in a specific way, but it has some issues with how it handles the GCD calculations and the stopping condition.

### Key Points to Consider:
1. **Sorting Issue**: The code sorts the array in a non-standard way, which might not be efficient and could lead to incorrect results.
2. **GCD Calculation**: The GCD calculation is done in a loop, which might not be optimal and could lead to incorrect results if the array is not properly sorted.
3. **Stopping Condition**: The stopping condition for the loop is based on whether any element in the array has a GCD greater than 1, which might not always be the correct condition to break the loop.

### Test Input to Expose the Bug:
We need to create a test case where the GCD calculations and the stopping condition might fail. A good candidate is an array where all elements are the same, or an array with a mix of small and large numbers that have a common factor.

### Example Test Input:
```plaintext
1
5
10 20 30 40 50
```

### Explanation:
- **Array**: [10, 20, 30, 40, 50]
- **Expected Output**: The smallest possible value of the expression is 15.
- **Reason**: The GCD of the entire array is 10. The optimal arrangement is [10, 20, 30, 40, 50], which results in:
  - \(\gcd(10) = 10\)
  - \(\gcd(10, 20) = 10\)
  - \(\gcd(10, 20, 30) = 10\)
  - \(\gcd(10, 20, 30, 40) = 10\)
  - \(\gcd(10, 20, 30, 40, 50) = 10\)
  - Sum = 10 + 10 + 10 + 10 + 10 = 50

However, the buggy code might not correctly handle this case because it sorts the array and then incorrectly computes the GCD, potentially leading to an incorrect result.

### Output:
```plaintext
1
5
10 20 30 40 50
```