**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alter the GCD
time_limit_ms: 4000
memory_limit_mb: 256
Description: You are given two arrays $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$.

You must perform the following operation exactly once:

  * choose any indices $l$ and $r$ such that $1 \le l \le r \le n$;    * swap $a_i$ and $b_i$ for all $i$ such that $l \leq i \leq r$.

Find the maximum possible value of $\text{gcd}(a_1, a_2, \ldots, a_n) + \text{gcd}(b_1, b_2, \ldots, b_n)$ after performing the operation exactly once. Also find the number of distinct pairs $(l, r)$ which achieve the maximum value.

Input: In the first line of the input, you are given a single integer $t$ ($1 \le t \le 10^5$), the number of test cases. Then the description of each test case follows.

In the first line of each test case, you are given a single integer $n$ ($1 \le n \le 2 \cdot 10^5$), representing the number of integers in each array.

In the next line, you are given $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the array $a$.

In the last line, you are given $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 10^9$) — the elements of the array $b$.

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

Output: For each test case, output a line with two integers: the maximum value of $\text{gcd}(a_1, a_2, \ldots, a_n) + \text{gcd}(b_1, b_2, \ldots, b_n)$ after performing the operation exactly once, and the number of ways.

Examples:
input:
5
8
11 4 16 17 3 24 25 8
8 10 4 21 17 18 25 21
4
6 4 24 13
15 3 1 14
2
13 14
5 8
8
20 17 15 11 21 10 3 7
9 9 4 20 14 9 13 1
2
18 13
15 20
output:
2 36
3 2
2 3
2 36
6 1

Note: In the first, third, and fourth test cases, there's no way to achieve a higher GCD than $1$ in any of the arrays, so the answer is $1 + 1 = 2$. Any pair $(l, r)$ achieves the same result; for example, in the first test case there are $36$ such pairs.

In the last test case, you must choose $l = 1$, $r = 2$ to maximize the answer. In this case, the GCD of the first array is $5$, and the GCD of the second array is $1$, so the answer is $5 + 1 = 6$, and the number of ways is $1$.

Buggy Code:
import java.util.*;

public class Main {
    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];
            int[] b = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
            }
            for (int i = 0; i < n; i++) {
                b[i] = sc.nextInt();
            }

            Map<String, Long> dp = new HashMap<>();
            String key1 = 0 + "," + a[0] + "," + b[0];
            String key2 = 1 + "," + b[0] + "," + a[0];
            String key3 = 2 + "," + b[0] + "," + a[0];

            dp.put(key1, 1L);
            dp.put(key2, 1L);
            dp.put(key3, 1L);

            for (int i = 1; i < n; i++) {
                Map<String, Long> dp2 = new HashMap<>();
                int ai = a[i], bi = b[i];
                for (Map.Entry<String, Long> entry : dp.entrySet()) {
                    String[] parts = entry.getKey().split(",");
                    int ph = Integer.parseInt(parts[0]);
                    int g1 = Integer.parseInt(parts[1]);
                    int g2 = Integer.parseInt(parts[2]);
                    long f = entry.getValue();

                    if (ph == 2) {
                        String newKey = 2 + "," + gcd(ai, g1) + "," + gcd(bi, g2);
                        dp2.put(newKey, dp2.getOrDefault(newKey, 0L) + f);
                        continue;
                    }
                    if (ph == 1) {
                        String newKey1 = 1 + "," + gcd(bi, g1) + "," + gcd(ai, g2);
                        String newKey2 = 2 + "," + gcd(bi, g1) + "," + gcd(ai, g2);
                        dp2.put(newKey1, dp2.getOrDefault(newKey1, 0L) + f);
                        dp2.put(newKey2, dp2.getOrDefault(newKey2, 0L) + f);
                        continue;
                    }
                    String newKey0 = 0 + "," + gcd(ai, g1) + "," + gcd(bi, g2);
                    String newKey1 = 1 + "," + gcd(bi, g1) + "," + gcd(ai, g2);
                    String newKey2 = 2 + "," + gcd(bi, g1) + "," + gcd(ai, g2);
                    dp2.put(newKey0, dp2.getOrDefault(newKey0, 0L) + f);
                    dp2.put(newKey1, dp2.getOrDefault(newKey1, 0L) + f);
                    dp2.put(newKey2, dp2.getOrDefault(newKey2, 0L) + f);
                }
                dp = dp2;
            }
            Map<Integer, Long> res = new HashMap<>();
            for (Map.Entry<String, Long> entry : dp.entrySet()) {
                String[] parts = entry.getKey().split(",");
                int ph = Integer.parseInt(parts[0]);
                int g1 = Integer.parseInt(parts[1]);
                int g2 = Integer.parseInt(parts[2]);
                long f = entry.getValue();
                if (ph != 2) continue;
                int sum = g1 + g2;
                res.put(sum, res.getOrDefault(sum, 0L) + f);
            }
            int bestSum = Collections.max(res.keySet());
            long bestCount = res.get(bestSum);
            System.out.println(bestSum + " " + bestCount);
        }
    }
    private static int gcd(int a, int b) {
        while (b != 0) {
            int temp = b;
            b = a % b;
            a = 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:
```plaintext
1
3
1 2 3
4 5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Circuit
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has just crafted a circuit with $n$ lights and $2n$ switches. Each component (a light or a switch) has two states: on or off. The lights and switches are arranged in a way that:

  * Each light is connected to exactly two switches.    * Each switch is connected to exactly one light. It's unknown which light each switch is connected to.    * When all switches are off, all lights are also off.    * If a switch is toggled (from on to off, or vice versa), the state of the light connected to it will also toggle.

Alice brings the circuit, which shows only the states of the $2n$ switches, to her sister Iris and gives her a riddle: what is the minimum and maximum number of lights that can be turned on?

Knowing her little sister's antics too well, Iris takes no more than a second to give Alice a correct answer. Can you do the same?

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 50$) — the number of lights in the circuit.

The second line of each test case contains $2n$ integers $a_1, a_2, \ldots, a_{2n}$ ($0 \le a_i \le 1$) — the states of the switches in the circuit. $a_i = 0$ means the $i$-th switch is off, and $a_i = 1$ means the $i$-th switch is on.

Output: For each test case, output two integers — the minimum and maximum number of lights, respectively, that can be turned on.

Examples:
input:
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
output:
0 0
1 1
0 0
0 2
1 3

Note: In the first test case, there is only one light in the circuit, and no switch is on, so the light is certainly off.

In the second test case, there is only one light in the circuit, but one switch connected to it is on, so the light is on.

In the third test case, there is only one light in the circuit, and both switches are on, so the light is off as it was toggled twice.

In the fourth test case, to have no lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $4$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $2$ and switch $6$ are connected to light $2$. Since both switches are off, light $2$ is also off.    * Switch $3$ and switch $5$ are connected to light $3$. Both switches are on, so light $3$ is toggled twice from its initial off state, and thus also stays off.

And to have $2$ lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $2$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $3$ and switch $4$ are connected to light $2$. Since switch $3$ is on and switch $4$ is off, light $2$ is toggled once from its initial off state, so it is on.    * Switch $5$ and switch $6$ are connected to light $3$. Since switch $5$ is on and switch $6$ is off, light $3$ is toggled once from its initial off state, so it is on.

Buggy Code:
/*// Ih Badhon Chowdhury //*/

#include <bits/stdc++.h>
#include<ext/pb_ds/tree_policy.hpp>
#include<ext/pb_ds/assoc_container.hpp>
using namespace std;

//using namespace __gnu_pbds;
//template<typename T> using orderset = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
//template<typename T> using ordermultiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
//(*X).find_by_order(index) /// return value (0 index)
//(X).order_of_key(value) /// return lower_bound(value)

/*
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); /// mt19937_64 (long long)
auto my_rand(long long l,long long r)
{
  return uniform_int_distribution<long long>(l,r)(rng);
}
#ifndef ONLINE_JUDGE
#include "debug.h"
#else
#define dbg(...) 42
#endif
*/

#define  ll        long long int
#define  ld        long double
#define  sp        " "
#define  lcm(a,b)  (a*b)/__gcd(a,b)
#define  cinv(v)   for(auto &i:v) cin >> i;
#define  vi        vector<int>
#define  vii       vector<ll>
#define  coutv(v)  for(auto it:v)cout<<it<<' ';cout<<endl;
#define  yes       cout << "YES" << endl
#define  no        cout << "NO" << endl
#define  case      "Case " << test++ << ": "
#define  all(v)    v.begin(), v.end()
#define  Faster    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define  nl        '\n'
#define PI acos(-1)

const int N = 1000001;
const int mod = 1000000007;
const ll INF = LLONG_MAX;

///.........Bit_Manipulation...........///
#define MSB(mask) 63-__builtin_clzll(mask)  /// 0 -> -1
#define LSB(mask) __builtin_ctzll(mask)  /// 0 -> 64
#define ONE(mask) __builtin_popcountll(mask)
#define CHECK(mask,bit) (mask&(1LL<<bit))
#define ON(mask,bit) (mask(1LL<<bit))
#define OFF(mask,bit) (mask&~(1LL<<bit))
#define CHANGE(mask,bit) (mask^(1LL<<bit))
#define LOG2(x) (__builtin_clz(x) ^ 31)
#define LSB1(mask) (__builtin_ffs(mask) - 1)
#define STRING_TO_INT(mask) (std::bitset<64>(mask).to_ullong())

///................Graph's Move.................
///const int dx[] = {+1,-1,+0,+0}; ///Rock's Move
///const int dy[] = {+0,+0,+1,-1}; ///Rock's Move
///const int dx[] = {+0,+0,+1,-1,-1,+1,-1,+1}; ///King's Move
///const int dy[] = {-1,+1,+0,+0,+1,+1,-1,-1}; ///king's Move
///const int dx[] = {-2,-2,-1,-1,+1,+1,+2,+2}; ///knight's Move
///const int dy[] = {-1,+1,-2,+2,-2,+2,-1,+1}; ///knight's Move
///*.....................-_-........................*///

inline ll IPOW(ll n,ll k){ll ans=1;while(k){if(k&1)ans=ans*n%mod;n=n*n%mod;k>>=1;}return ans;}
inline ll POW(ll n,ll k){ll ans=1;while(k){if(k&1)ans=ans*n;n=n*n;k>>=1;}return ans;}
//////

// Driver Code
int main()
{
    Faster;
    ll tc = 1, test = 1;
    cin >> tc;
    while(tc--)
    {
        int n;
        cin >> n;
        vi v(2*n);
        cinv(v);

        int off = 0, on = 0;

        for(auto e:v)
        if(e==0) off++;
        else on++;

        int ans = min(n,on);

        if(on>n and (on-n)%2==1) ans--;

        cout << on%2 << sp << ans << nl;
    }
}

**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
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Circuit
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has just crafted a circuit with $n$ lights and $2n$ switches. Each component (a light or a switch) has two states: on or off. The lights and switches are arranged in a way that:

  * Each light is connected to exactly two switches.    * Each switch is connected to exactly one light. It's unknown which light each switch is connected to.    * When all switches are off, all lights are also off.    * If a switch is toggled (from on to off, or vice versa), the state of the light connected to it will also toggle.

Alice brings the circuit, which shows only the states of the $2n$ switches, to her sister Iris and gives her a riddle: what is the minimum and maximum number of lights that can be turned on?

Knowing her little sister's antics too well, Iris takes no more than a second to give Alice a correct answer. Can you do the same?

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 50$) — the number of lights in the circuit.

The second line of each test case contains $2n$ integers $a_1, a_2, \ldots, a_{2n}$ ($0 \le a_i \le 1$) — the states of the switches in the circuit. $a_i = 0$ means the $i$-th switch is off, and $a_i = 1$ means the $i$-th switch is on.

Output: For each test case, output two integers — the minimum and maximum number of lights, respectively, that can be turned on.

Examples:
input:
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
output:
0 0
1 1
0 0
0 2
1 3

Note: In the first test case, there is only one light in the circuit, and no switch is on, so the light is certainly off.

In the second test case, there is only one light in the circuit, but one switch connected to it is on, so the light is on.

In the third test case, there is only one light in the circuit, and both switches are on, so the light is off as it was toggled twice.

In the fourth test case, to have no lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $4$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $2$ and switch $6$ are connected to light $2$. Since both switches are off, light $2$ is also off.    * Switch $3$ and switch $5$ are connected to light $3$. Both switches are on, so light $3$ is toggled twice from its initial off state, and thus also stays off.

And to have $2$ lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $2$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $3$ and switch $4$ are connected to light $2$. Since switch $3$ is on and switch $4$ is off, light $2$ is toggled once from its initial off state, so it is on.    * Switch $5$ and switch $6$ are connected to light $3$. Since switch $5$ is on and switch $6$ is off, light $3$ is toggled once from its initial off state, so it is on.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define v vector<int>
#define f(i, k, n) for (int i = k; i < n; i++)
#define pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define go(a, x) for (auto &a : x)
#define yn(a) cout << ((a) ? "yes\n" : "no\n")
#define ll long long
#define ld long double
#define pii pair<int, int>
#define pll pair<ll, ll>
#define mii map<int, int>
#define umii unordered_map<int, int>
#define si set<int>
#define usi unordered_set<int>
#define msi multiset<int>
const int mod = 1e9 + 7;

int xorify(int a, int b)
{
    return a ^ b;
}
int andify(int a, int b)
{
    return a & b;
}

int add(int a, int b)
{
    return a + b;
}

int mul(int a, int b)
{
    return a * b;
}

bool fin(int x, msi &s)
{
    return s.find(x) != s.end();
}

void run(int args)
{
    int n;
    cin >> n;

    v a(2*n);
    int nulls = 0;
    int ones = 0;
    f(i, 0, 2*n)
    {
        cin >> a[i];
        if(a[i])
        {
            ones++;
        }
        else
        {
            nulls++;
        }
    }

    if(nulls > n || nulls == 0)
    {
        cout << "0 ";
    }
    else if(nulls == n)
    {
        cout << "1 ";
    }
    else if(nulls < n)
    {
        cout << n << " ";
    }

    cout << min(ones, nulls) << "\n";

}

int32_t main()
{
    int t;
    cin >> t;
    while (t--)
    {
        run(0);
    }
    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
3
1
1 0
2
0 0 1 1
3
0 1 1 0 1 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Circuit
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has just crafted a circuit with $n$ lights and $2n$ switches. Each component (a light or a switch) has two states: on or off. The lights and switches are arranged in a way that:

  * Each light is connected to exactly two switches.    * Each switch is connected to exactly one light. It's unknown which light each switch is connected to.    * When all switches are off, all lights are also off.    * If a switch is toggled (from on to off, or vice versa), the state of the light connected to it will also toggle.

Alice brings the circuit, which shows only the states of the $2n$ switches, to her sister Iris and gives her a riddle: what is the minimum and maximum number of lights that can be turned on?

Knowing her little sister's antics too well, Iris takes no more than a second to give Alice a correct answer. Can you do the same?

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 50$) — the number of lights in the circuit.

The second line of each test case contains $2n$ integers $a_1, a_2, \ldots, a_{2n}$ ($0 \le a_i \le 1$) — the states of the switches in the circuit. $a_i = 0$ means the $i$-th switch is off, and $a_i = 1$ means the $i$-th switch is on.

Output: For each test case, output two integers — the minimum and maximum number of lights, respectively, that can be turned on.

Examples:
input:
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
output:
0 0
1 1
0 0
0 2
1 3

Note: In the first test case, there is only one light in the circuit, and no switch is on, so the light is certainly off.

In the second test case, there is only one light in the circuit, but one switch connected to it is on, so the light is on.

In the third test case, there is only one light in the circuit, and both switches are on, so the light is off as it was toggled twice.

In the fourth test case, to have no lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $4$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $2$ and switch $6$ are connected to light $2$. Since both switches are off, light $2$ is also off.    * Switch $3$ and switch $5$ are connected to light $3$. Both switches are on, so light $3$ is toggled twice from its initial off state, and thus also stays off.

And to have $2$ lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $2$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $3$ and switch $4$ are connected to light $2$. Since switch $3$ is on and switch $4$ is off, light $2$ is toggled once from its initial off state, so it is on.    * Switch $5$ and switch $6$ are connected to light $3$. Since switch $5$ is on and switch $6$ is off, light $3$ is toggled once from its initial off state, so it is on.

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
#define lli long long int
#define vi vector<int>
#define vl vector<ll>
#define nl cout<<"\n";
#define no cout<<"NO\n"
#define yes cout<<"YES\n"

#define inp(v,n) for(int i=0;i<n;i++) cin>>v[i];
#define out(v,n) for(int i=0;i<n;i++) cout<<v[i]<<" ";

using namespace std;

int b_s(const vector<int>& a, int t) {
    int l = -1, r = a.size();
    while (r - l > 1) {
        int m = (l + r) / 2;
        if (a[m] < t) {
            l = m;
        } else {
            r = m;
        }
    }
    return l; // return the last index where a[l] < t
}

//it is used for the interactive problems
bool ask(int x,int y)
{
    cout<<"? "<<x<<" "<<y<<"\n";
    string s;
    cin>>s;
    return (s=="YES");
    // this is ask function
}

bool isprime[1000004];

void sieve()
{
    for(int i=2;i<1e6;i++)
    {
        isprime[i]=true;
    }
    for(int i=2;i<=1e6;i++)
    {
        if(isprime[i])
        {
            for(int j=i*i;j<=1e6;j+=i)
            {
                isprime[j]=false;
            }
        }
    }
}
void solve()
{
    ll n; cin >> n;
    n=2*n;
    vl a(n);
    inp(a, n);

    int zero=0,one=0;
    for(int i=0;i<n;i++)
    {
        if(a[i]==0)
        {
            zero++;
        }
        else{
            one++;
        }
    }
    int mini=0,maxi=0;
    if(zero==n||one==n)
    {
        cout<<0<<" "<<0;nl;
        return;
    }
    if(one%2!=0)
    {
        cout<<1<<" "<<one;nl;
        return;
    }
    cout<<0<<" "<<one;nl;

    // out(a,n);nl;
}

int main() {
    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 buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code has issues with handling cases where the number of switches in the "on" state is even, and it incorrectly calculates the minimum and maximum number of lights that can be turned on.

### Step-by-Step Analysis:

1. **Edge Case: All Switches Off or All Switches On**
   - The code correctly handles the case where all switches are off or all switches are on by outputting `0 0`.
   - This case is already covered and does not expose the bug.

2. **Odd Number of Switches On**
   - The code correctly handles the case where the number of switches in the "on" state is odd by outputting `1 {one}`.
   - This case is also correctly handled and does not expose the bug.

3. **Even Number of Switches On**
   - The code incorrectly handles the case where the number of switches in the "on" state is even. It always outputs `0 {one}`.
   - This is incorrect because the minimum number of lights that can be turned on can be more than 0 in some cases.

### Challenging Test Case:
To expose the bug, we need a test case where the number of switches in the "on" state is even, and the minimum number of lights that can be turned on is not 0. Specifically, we need a case where the switches can be arranged such that some lights are on, but not all.

### Test Case:
```plaintext
1
3
1 1 0 1 0 1
```

### Explanation:
- **Number of Lights (n):** 3
- **Switch States:** [1, 1, 0, 1, 0, 1]
- **Number of Switches On:** 4 (even)

### Correct Output:
- **Minimum Number of Lights On:** 2
- **Maximum Number of Lights On:** 3

### Reasoning:
- **Minimum Number of Lights On:** 2
  - Possible arrangement: Switch 1 and Switch 3 are connected to Light 1 (both on, so Light 1 is off).
  - Switch 2 and Switch 5 are connected to Light 2 (one on, one off, so Light 2 is on).
  - Switch 4 and Switch 6 are connected to Light 3 (one on, one off, so Light 3 is on).
- **Maximum Number of Lights On:** 3
  - Possible arrangement: Switch 1 and Switch 2 are connected to Light 1 (both on, so Light 1 is off).
  - Switch 3 and Switch 4 are connected to Light 2 (one on, one off, so Light 2 is on).
  - Switch 5 and Switch 6 are connected to Light 3 (one on, one off, so Light 3 is on).

The correct output should be `2 3`, but the buggy code will output `0 4`, which is incorrect.

### Test Input:
```plaintext
1
3
1 1 0 1 0 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Circuit
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has just crafted a circuit with $n$ lights and $2n$ switches. Each component (a light or a switch) has two states: on or off. The lights and switches are arranged in a way that:

  * Each light is connected to exactly two switches.    * Each switch is connected to exactly one light. It's unknown which light each switch is connected to.    * When all switches are off, all lights are also off.    * If a switch is toggled (from on to off, or vice versa), the state of the light connected to it will also toggle.

Alice brings the circuit, which shows only the states of the $2n$ switches, to her sister Iris and gives her a riddle: what is the minimum and maximum number of lights that can be turned on?

Knowing her little sister's antics too well, Iris takes no more than a second to give Alice a correct answer. Can you do the same?

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 50$) — the number of lights in the circuit.

The second line of each test case contains $2n$ integers $a_1, a_2, \ldots, a_{2n}$ ($0 \le a_i \le 1$) — the states of the switches in the circuit. $a_i = 0$ means the $i$-th switch is off, and $a_i = 1$ means the $i$-th switch is on.

Output: For each test case, output two integers — the minimum and maximum number of lights, respectively, that can be turned on.

Examples:
input:
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
output:
0 0
1 1
0 0
0 2
1 3

Note: In the first test case, there is only one light in the circuit, and no switch is on, so the light is certainly off.

In the second test case, there is only one light in the circuit, but one switch connected to it is on, so the light is on.

In the third test case, there is only one light in the circuit, and both switches are on, so the light is off as it was toggled twice.

In the fourth test case, to have no lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $4$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $2$ and switch $6$ are connected to light $2$. Since both switches are off, light $2$ is also off.    * Switch $3$ and switch $5$ are connected to light $3$. Both switches are on, so light $3$ is toggled twice from its initial off state, and thus also stays off.

And to have $2$ lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $2$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $3$ and switch $4$ are connected to light $2$. Since switch $3$ is on and switch $4$ is off, light $2$ is toggled once from its initial off state, so it is on.    * Switch $5$ and switch $6$ are connected to light $3$. Since switch $5$ is on and switch $6$ is off, light $3$ is toggled once from its initial off state, so it is on.

Buggy Code:
n = int(input())

for i in range(n):
    zeros = 0
    ones = 0
    l = int(input())
    s = input()
    zeros = s.count('0')
    ones = s.count('1')
    if zeros == 0 or ones == 0:
        print('0 0')
    elif zeros > ones:
        if ones % 2 != 0:
            print('1' + ' ' + str(ones))
        else:
            print('0' + ' ' + str(ones))
    elif ones > zeros:
        if zeros % 2 != 0:
            print('1' + ' ' + str(zeros))
        else:
            print('0' + ' ' + str(zeros))
    else:
        print('1' + ' ' + str(ones))

**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 maximize the failure probability. The buggy code has issues with handling cases where the number of zeros and ones is equal, and it also incorrectly handles cases where the number of zeros and ones is not equal.

Let's consider the following scenarios:

1. **All switches are off or all switches are on**: This is already handled correctly by the buggy code, so it's not a good candidate for exposing the bug.
2. **Equal number of zeros and ones**: This is a critical case where the buggy code might fail.
3. **More zeros than ones or more ones than zeros**: The buggy code has issues with the parity of the counts.

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

```plaintext
3
2
0 0 1 1
2
1 1 0 0
3
1 0 1 0 1 0
```

### Explanation:

1. **Test Case 1:**
   - **n = 2**
   - **Switch states: 0 0 1 1**
   - **Expected Output: 0 2**
     - Minimum: 0 (all lights can be off if both pairs of switches are off and on, respectively)
     - Maximum: 2 (both lights can be on if each light has one switch on and one switch off)

2. **Test Case 2:**
   - **n = 2**
   - **Switch states: 1 1 0 0**
   - **Expected Output: 0 2**
     - Minimum: 0 (all lights can be off if both pairs of switches are on and off, respectively)
     - Maximum: 2 (both lights can be on if each light has one switch on and one switch off)

3. **Test Case 3:**
   - **n = 3**
   - **Switch states: 1 0 1 0 1 0**
   - **Expected Output: 1 3**
     - Minimum: 1 (at least one light can be on if one light has both switches off, and the other two lights have one switch on and one switch off)
     - Maximum: 3 (all three lights can be on if each light has one switch on and one switch off)

These test cases are designed to challenge the buggy code's handling of equal and unequal counts of zeros and ones, as well as the parity of these counts.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Circuit
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has just crafted a circuit with $n$ lights and $2n$ switches. Each component (a light or a switch) has two states: on or off. The lights and switches are arranged in a way that:

  * Each light is connected to exactly two switches.    * Each switch is connected to exactly one light. It's unknown which light each switch is connected to.    * When all switches are off, all lights are also off.    * If a switch is toggled (from on to off, or vice versa), the state of the light connected to it will also toggle.

Alice brings the circuit, which shows only the states of the $2n$ switches, to her sister Iris and gives her a riddle: what is the minimum and maximum number of lights that can be turned on?

Knowing her little sister's antics too well, Iris takes no more than a second to give Alice a correct answer. Can you do the same?

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 50$) — the number of lights in the circuit.

The second line of each test case contains $2n$ integers $a_1, a_2, \ldots, a_{2n}$ ($0 \le a_i \le 1$) — the states of the switches in the circuit. $a_i = 0$ means the $i$-th switch is off, and $a_i = 1$ means the $i$-th switch is on.

Output: For each test case, output two integers — the minimum and maximum number of lights, respectively, that can be turned on.

Examples:
input:
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
output:
0 0
1 1
0 0
0 2
1 3

Note: In the first test case, there is only one light in the circuit, and no switch is on, so the light is certainly off.

In the second test case, there is only one light in the circuit, but one switch connected to it is on, so the light is on.

In the third test case, there is only one light in the circuit, and both switches are on, so the light is off as it was toggled twice.

In the fourth test case, to have no lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $4$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $2$ and switch $6$ are connected to light $2$. Since both switches are off, light $2$ is also off.    * Switch $3$ and switch $5$ are connected to light $3$. Both switches are on, so light $3$ is toggled twice from its initial off state, and thus also stays off.

And to have $2$ lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $2$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $3$ and switch $4$ are connected to light $2$. Since switch $3$ is on and switch $4$ is off, light $2$ is toggled once from its initial off state, so it is on.    * Switch $5$ and switch $6$ are connected to light $3$. Since switch $5$ is on and switch $6$ is off, light $3$ is toggled once from its initial off state, so it is on.

Buggy Code:
def solve(x,y):
    count1=0
    count0=0
    for i in y:
        if i==0:
            count0+=1
        else:
            count1+=1
    if count1==0 or count0==0:
        return [0,0]
    elif count1==2*x-1 or count0==2*x-1:
        return [1,1]
    elif max(count1,count0)//min(count1,count0)==2:
        return [0,x-1]
    else:
        return [1,x]

n=int(input())
for i in range(n):
    x=int(input())
    y=list(map(int,input().split()))
    z=solve(x,y)
    print(*z)

**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
2
0 0 0 0
2
1 1 1 1
2
1 0 1 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Circuit
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has just crafted a circuit with $n$ lights and $2n$ switches. Each component (a light or a switch) has two states: on or off. The lights and switches are arranged in a way that:

  * Each light is connected to exactly two switches.    * Each switch is connected to exactly one light. It's unknown which light each switch is connected to.    * When all switches are off, all lights are also off.    * If a switch is toggled (from on to off, or vice versa), the state of the light connected to it will also toggle.

Alice brings the circuit, which shows only the states of the $2n$ switches, to her sister Iris and gives her a riddle: what is the minimum and maximum number of lights that can be turned on?

Knowing her little sister's antics too well, Iris takes no more than a second to give Alice a correct answer. Can you do the same?

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 50$) — the number of lights in the circuit.

The second line of each test case contains $2n$ integers $a_1, a_2, \ldots, a_{2n}$ ($0 \le a_i \le 1$) — the states of the switches in the circuit. $a_i = 0$ means the $i$-th switch is off, and $a_i = 1$ means the $i$-th switch is on.

Output: For each test case, output two integers — the minimum and maximum number of lights, respectively, that can be turned on.

Examples:
input:
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
output:
0 0
1 1
0 0
0 2
1 3

Note: In the first test case, there is only one light in the circuit, and no switch is on, so the light is certainly off.

In the second test case, there is only one light in the circuit, but one switch connected to it is on, so the light is on.

In the third test case, there is only one light in the circuit, and both switches are on, so the light is off as it was toggled twice.

In the fourth test case, to have no lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $4$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $2$ and switch $6$ are connected to light $2$. Since both switches are off, light $2$ is also off.    * Switch $3$ and switch $5$ are connected to light $3$. Both switches are on, so light $3$ is toggled twice from its initial off state, and thus also stays off.

And to have $2$ lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $2$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $3$ and switch $4$ are connected to light $2$. Since switch $3$ is on and switch $4$ is off, light $2$ is toggled once from its initial off state, so it is on.    * Switch $5$ and switch $6$ are connected to light $3$. Since switch $5$ is on and switch $6$ is off, light $3$ is toggled once from its initial off state, so it is on.

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    arr = list(map(int,input().split()))
    ones = arr.count(1)
    zeros = arr.count(0)
    if ones % 2 == 0:
        mn = 0
        if zeros > 0 and ones > 0:
            max = ones
        else:
            max = 0
    else:
        mn = 1
        max = min(ones,zeros)
    print(mn, max)

**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
0 1
2
0 1 1 0
3
0 1 1 1 0 0
4
0 1 1 1 1 0 0 0
5
0 1 1 1 1 1 0 0 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Circuit
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has just crafted a circuit with $n$ lights and $2n$ switches. Each component (a light or a switch) has two states: on or off. The lights and switches are arranged in a way that:

  * Each light is connected to exactly two switches.    * Each switch is connected to exactly one light. It's unknown which light each switch is connected to.    * When all switches are off, all lights are also off.    * If a switch is toggled (from on to off, or vice versa), the state of the light connected to it will also toggle.

Alice brings the circuit, which shows only the states of the $2n$ switches, to her sister Iris and gives her a riddle: what is the minimum and maximum number of lights that can be turned on?

Knowing her little sister's antics too well, Iris takes no more than a second to give Alice a correct answer. Can you do the same?

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 50$) — the number of lights in the circuit.

The second line of each test case contains $2n$ integers $a_1, a_2, \ldots, a_{2n}$ ($0 \le a_i \le 1$) — the states of the switches in the circuit. $a_i = 0$ means the $i$-th switch is off, and $a_i = 1$ means the $i$-th switch is on.

Output: For each test case, output two integers — the minimum and maximum number of lights, respectively, that can be turned on.

Examples:
input:
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
output:
0 0
1 1
0 0
0 2
1 3

Note: In the first test case, there is only one light in the circuit, and no switch is on, so the light is certainly off.

In the second test case, there is only one light in the circuit, but one switch connected to it is on, so the light is on.

In the third test case, there is only one light in the circuit, and both switches are on, so the light is off as it was toggled twice.

In the fourth test case, to have no lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $4$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $2$ and switch $6$ are connected to light $2$. Since both switches are off, light $2$ is also off.    * Switch $3$ and switch $5$ are connected to light $3$. Both switches are on, so light $3$ is toggled twice from its initial off state, and thus also stays off.

And to have $2$ lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $2$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $3$ and switch $4$ are connected to light $2$. Since switch $3$ is on and switch $4$ is off, light $2$ is toggled once from its initial off state, so it is on.    * Switch $5$ and switch $6$ are connected to light $3$. Since switch $5$ is on and switch $6$ is off, light $3$ is toggled once from its initial off state, so it is on.

Buggy Code:
import java.util.*;
public class Main{
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        for(int i=1;i<=n;i++){
            int lights=sc.nextInt();
            int[] arr=new int[2*lights];
            int count=0;
            for(int j=0;j<arr.length;j++){
                arr[j]=sc.nextInt();
                count=count+arr[j];
            }
            System.out.print(count%2+" ");
            if(count<=lights){
                System.out.println(count);
            }else{
                System.out.println(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
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
```

### Explanation:
1. **Case 1:**
   - **Input:** `1 0 0`
   - **Expected Output:** `0 0`
   - **Reason:** Both switches are off, so the light is off.

2. **Case 2:**
   - **Input:** `1 0 1`
   - **Expected Output:** `1 1`
   - **Reason:** One switch is on, so the light is on.

3. **Case 3:**
   - **Input:** `1 1 1`
   - **Expected Output:** `0 0`
   - **Reason:** Both switches are on, so the light is off (toggled twice).

4. **Case 4:**
   - **Input:** `3 0 0 1 0 1 0`
   - **Expected Output:** `0 2`
   - **Reason:** To have no lights on, arrange the switches so that each light is toggled an even number of times. To have 2 lights on, arrange the switches so that two lights are toggled once and one light is toggled twice.

5. **Case 5:**
   - **Input:** `3 0 1 1 1 0 0`
   - **Expected Output:** `1 3`
   - **Reason:** To have 1 light on, arrange the switches so that one light is toggled once and the other two lights are toggled twice. To have 3 lights on, arrange the switches so that each light is toggled once.

These test cases cover a variety of scenarios, including edge cases with small and large numbers of lights and different switch states, to expose potential bugs in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Circuit
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has just crafted a circuit with $n$ lights and $2n$ switches. Each component (a light or a switch) has two states: on or off. The lights and switches are arranged in a way that:

  * Each light is connected to exactly two switches.    * Each switch is connected to exactly one light. It's unknown which light each switch is connected to.    * When all switches are off, all lights are also off.    * If a switch is toggled (from on to off, or vice versa), the state of the light connected to it will also toggle.

Alice brings the circuit, which shows only the states of the $2n$ switches, to her sister Iris and gives her a riddle: what is the minimum and maximum number of lights that can be turned on?

Knowing her little sister's antics too well, Iris takes no more than a second to give Alice a correct answer. Can you do the same?

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 50$) — the number of lights in the circuit.

The second line of each test case contains $2n$ integers $a_1, a_2, \ldots, a_{2n}$ ($0 \le a_i \le 1$) — the states of the switches in the circuit. $a_i = 0$ means the $i$-th switch is off, and $a_i = 1$ means the $i$-th switch is on.

Output: For each test case, output two integers — the minimum and maximum number of lights, respectively, that can be turned on.

Examples:
input:
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
output:
0 0
1 1
0 0
0 2
1 3

Note: In the first test case, there is only one light in the circuit, and no switch is on, so the light is certainly off.

In the second test case, there is only one light in the circuit, but one switch connected to it is on, so the light is on.

In the third test case, there is only one light in the circuit, and both switches are on, so the light is off as it was toggled twice.

In the fourth test case, to have no lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $4$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $2$ and switch $6$ are connected to light $2$. Since both switches are off, light $2$ is also off.    * Switch $3$ and switch $5$ are connected to light $3$. Both switches are on, so light $3$ is toggled twice from its initial off state, and thus also stays off.

And to have $2$ lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $2$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $3$ and switch $4$ are connected to light $2$. Since switch $3$ is on and switch $4$ is off, light $2$ is toggled once from its initial off state, so it is on.    * Switch $5$ and switch $6$ are connected to light $3$. Since switch $5$ is on and switch $6$ is off, light $3$ is toggled once from its initial off state, so it is on.

Buggy Code:
import java.util.*;
public class Lights
{
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for(int k=0;k<t;k++)
        {
            int n = sc.nextInt();
            int s[] = new int[n*2];
            for(int l=0;l<n*2;l++)
            {
                s[l]=sc.nextInt();
            }
            int l = n, max=0;
            int g=0,min=n;
            for(int i=0;i<n*2;i++)
            {
                for(int j=i+1;j<n*2;j++)
                {
                    int res = s[i]^s[j];
                    if(res==1){
                    g++;}
                    if(res==0){
                    l--;}
                }
                if(g>max)
                 max = g;
                if(l<min)
                 min = l;
                g=0;l=n;
            }
            System.out.println(min + " " + max);
        }
        sc.close();
    }
}

**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
0 0
2
1 0 0 1
3
0 1 1 0 1 0
```

### Explanation:
1. **Test Case 1:**
   - **Input:** `1 0 0`
   - **Expected Output:** `0 0`
   - **Reasoning:** There is only one light, and both switches are off. Therefore, the light must be off.

2. **Test Case 2:**
   - **Input:** `2 1 0 0 1`
   - **Expected Output:** `1 1`
   - **Reasoning:** There are two lights, and the switches are arranged such that one light is on and the other is off. The minimum and maximum number of lights that can be on is 1.

3. **Test Case 3:**
   - **Input:** `3 0 1 1 0 1 0`
   - **Expected Output:** `1 3`
   - **Reasoning:** There are three lights, and the switches are arranged such that:
     - Light 1 can be off (both switches are off or both are on).
     - Light 2 and Light 3 can be on (one switch is on and the other is off).
     - The minimum number of lights that can be on is 1, and the maximum is 3.

These test cases are designed to cover different scenarios and edge cases, including the simplest case with one light and more complex cases with multiple lights and switches. They should help expose any bugs in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Circuit
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has just crafted a circuit with $n$ lights and $2n$ switches. Each component (a light or a switch) has two states: on or off. The lights and switches are arranged in a way that:

  * Each light is connected to exactly two switches.    * Each switch is connected to exactly one light. It's unknown which light each switch is connected to.    * When all switches are off, all lights are also off.    * If a switch is toggled (from on to off, or vice versa), the state of the light connected to it will also toggle.

Alice brings the circuit, which shows only the states of the $2n$ switches, to her sister Iris and gives her a riddle: what is the minimum and maximum number of lights that can be turned on?

Knowing her little sister's antics too well, Iris takes no more than a second to give Alice a correct answer. Can you do the same?

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 50$) — the number of lights in the circuit.

The second line of each test case contains $2n$ integers $a_1, a_2, \ldots, a_{2n}$ ($0 \le a_i \le 1$) — the states of the switches in the circuit. $a_i = 0$ means the $i$-th switch is off, and $a_i = 1$ means the $i$-th switch is on.

Output: For each test case, output two integers — the minimum and maximum number of lights, respectively, that can be turned on.

Examples:
input:
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
output:
0 0
1 1
0 0
0 2
1 3

Note: In the first test case, there is only one light in the circuit, and no switch is on, so the light is certainly off.

In the second test case, there is only one light in the circuit, but one switch connected to it is on, so the light is on.

In the third test case, there is only one light in the circuit, and both switches are on, so the light is off as it was toggled twice.

In the fourth test case, to have no lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $4$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $2$ and switch $6$ are connected to light $2$. Since both switches are off, light $2$ is also off.    * Switch $3$ and switch $5$ are connected to light $3$. Both switches are on, so light $3$ is toggled twice from its initial off state, and thus also stays off.

And to have $2$ lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $2$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $3$ and switch $4$ are connected to light $2$. Since switch $3$ is on and switch $4$ is off, light $2$ is toggled once from its initial off state, so it is on.    * Switch $5$ and switch $6$ are connected to light $3$. Since switch $5$ is on and switch $6$ is off, light $3$ is toggled once from its initial off state, so it is on.

Buggy Code:
// Kalp
#include<bits/stdc++.h>
using namespace std;
#define pi 3.14159265358979323846264338327950
#define fast_io ios_base::sync_with_stdio(false);cin.tie(NULL)
#define endl "\n"
#define int long long

#ifndef ONLINE_JUDGE
#define debug(x) cerr << #x << " "; _print(x); cerr << endl;
#else
#define debug(x);
#endif

typedef long long ll;
typedef unsigned long long ull;
typedef long double lld;
void _print(int t){cerr<<t;}
//void _print(ll t){cerr<<t;}
void _print(char t){cerr<<t;}
void _print(string t){cerr<<t;}
void _print(float t){cerr<<t;}
void _print(double t){cerr<<t;}
void _print(ull t){cerr<<t;}
template <class T, class V> void _print(pair <T, V> p);
template <class T, class V> void _print(map <T, V> mp);
template <class T> void _print(vector<T> v);
template <class T> void _print(set<T> s);
template <class T> void _print(multiset<T> s);
template <class T> void _print(T arr[]) {cerr << "[ "; for(T i: arr) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.first); cerr << ","; _print(p.second); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << endl;} cerr << "]";}
const int N = 1e7+7;
const int MOD = 1e9+7;
const ll INF = LLONG_MAX >> 1;
void init_code(){
   fast_io;
   #ifndef ONLINE_JUDGE
      freopen("input.txt", "r", stdin);
      freopen("output.txt", "w", stdout);
      freopen("error.txt", "w", stderr);
   #endif
}
// --------------------------------------------------------------------------------------------------------------------------------------//

void solve() {
   int n;
   cin>>n;
   int sz = 2*n;
   vector<int>arr(sz);
   int cnt = 0;
   for(int i=0; i<sz; i++) {
      cin>>arr[i];
      if(arr[i]==1) cnt++;
   }
   if(n==1) {
      if(cnt == 0) {
         cout<<0<<" "<<0<<endl;
      }
      else if(cnt == 1) {
         cout<<1<<" "<<1<<endl;
      }
      else cout<<0<<" "<<0<<endl;
      return;
   }
   debug(arr)
   int res = (cnt <= sz)? cnt: sz - (cnt%sz);
   cout<<cnt / n <<" "<< res << endl;

}

signed main() {
   init_code();
   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 create a test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that the current code might handle incorrectly. Let's analyze the code and identify potential issues:

### Code Analysis:
1. **Single Light Case (n = 1):**
   - The code handles this case correctly by checking the number of switches that are on.
   - If both switches are off, the light is off.
   - If one switch is on, the light is on.
   - If both switches are on, the light is off.

2. **Multiple Lights Case (n > 1):**
   - The code calculates the number of switches that are on (`cnt`).
   - It then calculates the minimum number of lights that can be on as `cnt / n`.
   - The maximum number of lights that can be on is calculated as `res = (cnt <= sz)? cnt: sz - (cnt % sz)`.

### Potential Bugs:
1. **Minimum Number of Lights Calculation:**
   - The minimum number of lights that can be on should be `0` if the number of switches that are on is even, and `1` if it is odd. This is because each light is toggled by two switches, and if the total number of toggles (on switches) is even, all lights will be in their initial state (off). If it is odd, one light will be on.
   - The current code uses `cnt / n`, which is incorrect. For example, if `cnt = 3` and `n = 2`, the minimum should be `1`, but the code will output `1`.

2. **Maximum Number of Lights Calculation:**
   - The maximum number of lights that can be on is the number of switches that are on divided by 2, because each light is toggled by two switches.
   - The current code uses a complex formula that might not always give the correct result. For example, if `cnt = 3` and `n = 2`, the maximum should be `1`, but the code will output `1`.

### Test Input:
To expose these bugs, we can use the following test input:

```plaintext
3
2
1 1 0 0
2
1 0 1 0
3
1 1 1 0 0 0
```

### Explanation:
1. **Test Case 1:**
   - `n = 2`, `a = [1, 1, 0, 0]`
   - `cnt = 2`
   - Expected output: `0 1` (since both switches being on will toggle one light, and the other light will remain off)
   - Current code output: `1 2` (incorrect)

2. **Test Case 2:**
   - `n = 2`, `a = [1, 0, 1, 0]`
   - `cnt = 2`
   - Expected output: `1 1` (since the toggles will result in one light being on and one light being off)
   - Current code output: `1 2` (incorrect)

3. **Test Case 3:**
   - `n = 3`, `a = [1, 1, 1, 0, 0, 0]`
   - `cnt = 3`
   - Expected output: `1 1` (since three toggles will result in one light being on and the other two lights being off)
   - Current code output: `1 3` (incorrect)

This test input should help expose the bugs in the code's logic for calculating the minimum and maximum number of lights that can be on.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Circuit
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has just crafted a circuit with $n$ lights and $2n$ switches. Each component (a light or a switch) has two states: on or off. The lights and switches are arranged in a way that:

  * Each light is connected to exactly two switches.    * Each switch is connected to exactly one light. It's unknown which light each switch is connected to.    * When all switches are off, all lights are also off.    * If a switch is toggled (from on to off, or vice versa), the state of the light connected to it will also toggle.

Alice brings the circuit, which shows only the states of the $2n$ switches, to her sister Iris and gives her a riddle: what is the minimum and maximum number of lights that can be turned on?

Knowing her little sister's antics too well, Iris takes no more than a second to give Alice a correct answer. Can you do the same?

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 50$) — the number of lights in the circuit.

The second line of each test case contains $2n$ integers $a_1, a_2, \ldots, a_{2n}$ ($0 \le a_i \le 1$) — the states of the switches in the circuit. $a_i = 0$ means the $i$-th switch is off, and $a_i = 1$ means the $i$-th switch is on.

Output: For each test case, output two integers — the minimum and maximum number of lights, respectively, that can be turned on.

Examples:
input:
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
output:
0 0
1 1
0 0
0 2
1 3

Note: In the first test case, there is only one light in the circuit, and no switch is on, so the light is certainly off.

In the second test case, there is only one light in the circuit, but one switch connected to it is on, so the light is on.

In the third test case, there is only one light in the circuit, and both switches are on, so the light is off as it was toggled twice.

In the fourth test case, to have no lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $4$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $2$ and switch $6$ are connected to light $2$. Since both switches are off, light $2$ is also off.    * Switch $3$ and switch $5$ are connected to light $3$. Both switches are on, so light $3$ is toggled twice from its initial off state, and thus also stays off.

And to have $2$ lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $2$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $3$ and switch $4$ are connected to light $2$. Since switch $3$ is on and switch $4$ is off, light $2$ is toggled once from its initial off state, so it is on.    * Switch $5$ and switch $6$ are connected to light $3$. Since switch $5$ is on and switch $6$ is off, light $3$ is toggled once from its initial off state, so it is on.

Buggy Code:
#include <iostream>

int main()
{
    short t;
    std::cin >> t;
    while (t--)
    {
        short n;
        std::cin >> n;
        short a[2 * n], count_one = 0;
        for (short i = 0; i < 2 * n; i++)
        {
            std::cin >> a[i];
            if (a[i] == 1)
            {
                count_one++;
            }
        }
        if (count_one >= 2 * n)
        {
            std::cout << count_one % 2 << " " << 2 * n - count_one << "\n";
        }
        else
        {
            std::cout << count_one % 2 << " " << count_one << "\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 provided buggy code, we need to focus on the logic that calculates the minimum and maximum number of lights that can be turned on. The buggy code has a critical error in its logic for determining the maximum number of lights that can be turned on. Specifically, it incorrectly handles the case where the number of switches that are on is less than the total number of switches.

Let's consider the following scenarios:

1. **All switches are off**: In this case, the minimum and maximum number of lights that can be turned on is 0.
2. **All switches are on**: In this case, the minimum number of lights that can be turned on is the number of lights modulo 2 (since toggling a light twice returns it to its original state), and the maximum number of lights that can be turned on is 0 (since all lights will be toggled back to off).
3. **Mixed states of switches**: This is where the buggy code fails. The code incorrectly calculates the maximum number of lights that can be turned on by simply using the count of switches that are on. However, the correct maximum number of lights that can be turned on should be the minimum of the number of on switches and the total number of lights.

To expose this bug, we can create a test case where the number of switches that are on is less than the total number of switches, and the number of lights is greater than the number of on switches. This will ensure that the buggy code's logic for the maximum number of lights will fail.

Here's a small-scale test input that meets these criteria:

```plaintext
2
3
0 0 0 1 1 0
4
0 0 0 0 1 1 1 0
```

Explanation:
- In the first test case:
  - There are 3 lights and 6 switches.
  - The switches are in the states [0, 0, 0, 1, 1, 0].
  - The number of switches that are on is 2.
  - The correct output should be:
    - Minimum: 0 (since the number of on switches is even, and all lights can be toggled back to off).
    - Maximum: 2 (since the maximum number of lights that can be turned on is the number of on switches).
- In the second test case:
  - There are 4 lights and 8 switches.
  - The switches are in the states [0, 0, 0, 0, 1, 1, 1, 0].
  - The number of switches that are on is 3.
  - The correct output should be:
    - Minimum: 1 (since the number of on switches is odd, and one light will be on).
    - Maximum: 3 (since the maximum number of lights that can be turned on is the number of on switches).

This test input should effectively expose the bug 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: Circuit
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has just crafted a circuit with $n$ lights and $2n$ switches. Each component (a light or a switch) has two states: on or off. The lights and switches are arranged in a way that:

  * Each light is connected to exactly two switches.    * Each switch is connected to exactly one light. It's unknown which light each switch is connected to.    * When all switches are off, all lights are also off.    * If a switch is toggled (from on to off, or vice versa), the state of the light connected to it will also toggle.

Alice brings the circuit, which shows only the states of the $2n$ switches, to her sister Iris and gives her a riddle: what is the minimum and maximum number of lights that can be turned on?

Knowing her little sister's antics too well, Iris takes no more than a second to give Alice a correct answer. Can you do the same?

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 50$) — the number of lights in the circuit.

The second line of each test case contains $2n$ integers $a_1, a_2, \ldots, a_{2n}$ ($0 \le a_i \le 1$) — the states of the switches in the circuit. $a_i = 0$ means the $i$-th switch is off, and $a_i = 1$ means the $i$-th switch is on.

Output: For each test case, output two integers — the minimum and maximum number of lights, respectively, that can be turned on.

Examples:
input:
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
output:
0 0
1 1
0 0
0 2
1 3

Note: In the first test case, there is only one light in the circuit, and no switch is on, so the light is certainly off.

In the second test case, there is only one light in the circuit, but one switch connected to it is on, so the light is on.

In the third test case, there is only one light in the circuit, and both switches are on, so the light is off as it was toggled twice.

In the fourth test case, to have no lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $4$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $2$ and switch $6$ are connected to light $2$. Since both switches are off, light $2$ is also off.    * Switch $3$ and switch $5$ are connected to light $3$. Both switches are on, so light $3$ is toggled twice from its initial off state, and thus also stays off.

And to have $2$ lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $2$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $3$ and switch $4$ are connected to light $2$. Since switch $3$ is on and switch $4$ is off, light $2$ is toggled once from its initial off state, so it is on.    * Switch $5$ and switch $6$ are connected to light $3$. Since switch $5$ is on and switch $6$ is off, light $3$ is toggled once from its initial off state, so it is on.

Buggy Code:
n=int(input())

for i in range(n):
    mi=0
    ma=0
    a=int(input())
    b=list(map(int,input().split()))
    for j in b:
        if j==1:

            ma+=1
            if ma>a:
                ma-=2
            if mi>0:
                mi-=1
            else:
                mi+=1
    print(mi,ma)

**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
0 0
1
0 1
1
1 1
2
0 0 1 1
2
1 0 1 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Circuit
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has just crafted a circuit with $n$ lights and $2n$ switches. Each component (a light or a switch) has two states: on or off. The lights and switches are arranged in a way that:

  * Each light is connected to exactly two switches.    * Each switch is connected to exactly one light. It's unknown which light each switch is connected to.    * When all switches are off, all lights are also off.    * If a switch is toggled (from on to off, or vice versa), the state of the light connected to it will also toggle.

Alice brings the circuit, which shows only the states of the $2n$ switches, to her sister Iris and gives her a riddle: what is the minimum and maximum number of lights that can be turned on?

Knowing her little sister's antics too well, Iris takes no more than a second to give Alice a correct answer. Can you do the same?

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 50$) — the number of lights in the circuit.

The second line of each test case contains $2n$ integers $a_1, a_2, \ldots, a_{2n}$ ($0 \le a_i \le 1$) — the states of the switches in the circuit. $a_i = 0$ means the $i$-th switch is off, and $a_i = 1$ means the $i$-th switch is on.

Output: For each test case, output two integers — the minimum and maximum number of lights, respectively, that can be turned on.

Examples:
input:
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
output:
0 0
1 1
0 0
0 2
1 3

Note: In the first test case, there is only one light in the circuit, and no switch is on, so the light is certainly off.

In the second test case, there is only one light in the circuit, but one switch connected to it is on, so the light is on.

In the third test case, there is only one light in the circuit, and both switches are on, so the light is off as it was toggled twice.

In the fourth test case, to have no lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $4$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $2$ and switch $6$ are connected to light $2$. Since both switches are off, light $2$ is also off.    * Switch $3$ and switch $5$ are connected to light $3$. Both switches are on, so light $3$ is toggled twice from its initial off state, and thus also stays off.

And to have $2$ lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $2$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $3$ and switch $4$ are connected to light $2$. Since switch $3$ is on and switch $4$ is off, light $2$ is toggled once from its initial off state, so it is on.    * Switch $5$ and switch $6$ are connected to light $3$. Since switch $5$ is on and switch $6$ is off, light $3$ is toggled once from its initial off state, so it is on.

Buggy Code:
# Author : Omm Praksh Sahoo
# Calm, Down... Wrong Submission Have 10 Mins Penalty

I = lambda: input()
II = lambda: int(input())
MII = lambda: map(int, input().split())
LI = lambda: list(input().split())
LII = lambda: list(map(int, input().split()))
LGMII = lambda: map(lambda x: int(x) - 1, input().split())
LGLII = lambda: list(map(lambda x: int(x) - 1, input().split()))

inf = float('inf')

#########################################################################################################
#-----------------------------------------------Code----------------------------------------------------#
#########################################################################################################
#

def solve():

    ans=0
    n1=II()
    l=LII()
    n=len(l)
    if n==2:
        if l.count(0)==2:
            return [0,0]
        elif l.count(1)==2:
            return [0,0]
        else:
            return [1,1]

    mi=0
    ma=0

    ma=min(l.count(1),n1)
    mi=ma
    # ans=[]

    for i in range (1,n-1):
        if l[i-1]==1 and l[i+1]==1:
            mi-=2
            mi=max(mi,0)
            i+=2
    # mi=l.count(1)

    return [mi,ma]

t=1
t=int(input())
for _ in range (t):
    print(*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
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Circuit
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has just crafted a circuit with $n$ lights and $2n$ switches. Each component (a light or a switch) has two states: on or off. The lights and switches are arranged in a way that:

  * Each light is connected to exactly two switches.    * Each switch is connected to exactly one light. It's unknown which light each switch is connected to.    * When all switches are off, all lights are also off.    * If a switch is toggled (from on to off, or vice versa), the state of the light connected to it will also toggle.

Alice brings the circuit, which shows only the states of the $2n$ switches, to her sister Iris and gives her a riddle: what is the minimum and maximum number of lights that can be turned on?

Knowing her little sister's antics too well, Iris takes no more than a second to give Alice a correct answer. Can you do the same?

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 50$) — the number of lights in the circuit.

The second line of each test case contains $2n$ integers $a_1, a_2, \ldots, a_{2n}$ ($0 \le a_i \le 1$) — the states of the switches in the circuit. $a_i = 0$ means the $i$-th switch is off, and $a_i = 1$ means the $i$-th switch is on.

Output: For each test case, output two integers — the minimum and maximum number of lights, respectively, that can be turned on.

Examples:
input:
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
output:
0 0
1 1
0 0
0 2
1 3

Note: In the first test case, there is only one light in the circuit, and no switch is on, so the light is certainly off.

In the second test case, there is only one light in the circuit, but one switch connected to it is on, so the light is on.

In the third test case, there is only one light in the circuit, and both switches are on, so the light is off as it was toggled twice.

In the fourth test case, to have no lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $4$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $2$ and switch $6$ are connected to light $2$. Since both switches are off, light $2$ is also off.    * Switch $3$ and switch $5$ are connected to light $3$. Both switches are on, so light $3$ is toggled twice from its initial off state, and thus also stays off.

And to have $2$ lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $2$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $3$ and switch $4$ are connected to light $2$. Since switch $3$ is on and switch $4$ is off, light $2$ is toggled once from its initial off state, so it is on.    * Switch $5$ and switch $6$ are connected to light $3$. Since switch $5$ is on and switch $6$ is off, light $3$ is toggled once from its initial off state, so it is on.

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

public class Main{

	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[2*n];
		    int on=0;
		    int of=0;
		    for(int i=0;i<2*n;i++){
		        a[i]=sc.nextInt();
		    }
		    for(int e:a){
		        if(e==1){
		            on++;
		        }
		        else{
		            of++;
		        }
		    }
		    int min=(of==0)?0:((on%2==0)?0:1);
		    int max=(of==0)?0:Math.min(n,on);
		    System.out.println(min+" "+max);
		}

	}
}

**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 where the logic in the code might fail. The key issues in the buggy code are:

1. The calculation of the minimum number of lights that can be turned on.
2. The calculation of the maximum number of lights that can be turned on.

### Analysis of the Buggy Code

1. **Minimum Number of Lights On**:
   - The code calculates the minimum number of lights that can be turned on as `0` if all switches are off or if the number of switches that are on is even. However, this logic is incorrect. For example, if there are an even number of switches on, it doesn't necessarily mean that the minimum number of lights that can be turned on is `0`.

2. **Maximum Number of Lights On**:
   - The code calculates the maximum number of lights that can be turned on as the minimum of `n` (the number of lights) and the number of switches that are on. This logic is also incorrect. For example, if there are more switches on than the number of lights, the maximum number of lights that can be turned on should be `n`.

### Challenging Test Input

To expose these issues, we can create a test case where the number of switches on is even and the number of switches off is also even. This will challenge the minimum number of lights calculation. Additionally, we can create a test case where the number of switches on is greater than the number of lights, which will challenge the maximum number of lights calculation.

### Test Case

Let's create a test case with `n = 3` and an even number of switches on and off:

```plaintext
1
3
0 1 0 1 0 1
```

### Explanation

- **Input**:
  - `n = 3` (3 lights and 6 switches)
  - Switch states: `0 1 0 1 0 1`

- **Expected Output**:
  - **Minimum Number of Lights On**: 1 (since the switches are arranged in pairs, and each pair has one switch on, at least one light will be on)
  - **Maximum Number of Lights On**: 3 (since there are 3 switches on, all 3 lights can be on if each light is connected to a pair of switches where one switch is on)

- **Buggy Code Output**:
  - **Minimum Number of Lights On**: 0 (incorrect)
  - **Maximum Number of Lights On**: 3 (correct)

This test case will expose the bug in the minimum number of lights calculation. The maximum number of lights calculation is correct in this case, but other test cases can be created to further challenge this.

### Final Test Input

```plaintext
1
3
0 1 0 1 0 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Circuit
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has just crafted a circuit with $n$ lights and $2n$ switches. Each component (a light or a switch) has two states: on or off. The lights and switches are arranged in a way that:

  * Each light is connected to exactly two switches.    * Each switch is connected to exactly one light. It's unknown which light each switch is connected to.    * When all switches are off, all lights are also off.    * If a switch is toggled (from on to off, or vice versa), the state of the light connected to it will also toggle.

Alice brings the circuit, which shows only the states of the $2n$ switches, to her sister Iris and gives her a riddle: what is the minimum and maximum number of lights that can be turned on?

Knowing her little sister's antics too well, Iris takes no more than a second to give Alice a correct answer. Can you do the same?

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 50$) — the number of lights in the circuit.

The second line of each test case contains $2n$ integers $a_1, a_2, \ldots, a_{2n}$ ($0 \le a_i \le 1$) — the states of the switches in the circuit. $a_i = 0$ means the $i$-th switch is off, and $a_i = 1$ means the $i$-th switch is on.

Output: For each test case, output two integers — the minimum and maximum number of lights, respectively, that can be turned on.

Examples:
input:
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
output:
0 0
1 1
0 0
0 2
1 3

Note: In the first test case, there is only one light in the circuit, and no switch is on, so the light is certainly off.

In the second test case, there is only one light in the circuit, but one switch connected to it is on, so the light is on.

In the third test case, there is only one light in the circuit, and both switches are on, so the light is off as it was toggled twice.

In the fourth test case, to have no lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $4$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $2$ and switch $6$ are connected to light $2$. Since both switches are off, light $2$ is also off.    * Switch $3$ and switch $5$ are connected to light $3$. Both switches are on, so light $3$ is toggled twice from its initial off state, and thus also stays off.

And to have $2$ lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $2$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $3$ and switch $4$ are connected to light $2$. Since switch $3$ is on and switch $4$ is off, light $2$ is toggled once from its initial off state, so it is on.    * Switch $5$ and switch $6$ are connected to light $3$. Since switch $5$ is on and switch $6$ is off, light $3$ is toggled once from its initial off state, so it is on.

Buggy Code:
import java.util.*;

public class Main {
    static class Node {
        int data;
        Node left, right;

        public Node(int item) {
            data = item;
            left = right = null;
        }
    }
    static Scanner in = new Scanner(System.in);

    public static void main(String[] args) {
//        int t = 1;
        int t = in.nextInt();
        while (t-- > 0) {
            solve();
        }
    }

//B-127
static void solve() {
    StringBuilder sb = new StringBuilder();

    int n = in.nextInt();
    int[] a = new int[2 * n];

    int total = 0;

    for (int i = 0; i < 2 * n; i++) {
        a[i] = in.nextInt();
        total += a[i];
    }
    if (n == 1 && total == 2) {
        sb.append("0 0\n");
        System.out.print(sb);
        return;
    }

    int minOn = (total % 2 == 0) ? 0 : 1;
    int maxOn = Math.min(n, total);

    sb.append(minOn).append(" ").append(maxOn).append("\n");
    System.out.print(sb);
}
}

**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
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Circuit
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has just crafted a circuit with $n$ lights and $2n$ switches. Each component (a light or a switch) has two states: on or off. The lights and switches are arranged in a way that:

  * Each light is connected to exactly two switches.    * Each switch is connected to exactly one light. It's unknown which light each switch is connected to.    * When all switches are off, all lights are also off.    * If a switch is toggled (from on to off, or vice versa), the state of the light connected to it will also toggle.

Alice brings the circuit, which shows only the states of the $2n$ switches, to her sister Iris and gives her a riddle: what is the minimum and maximum number of lights that can be turned on?

Knowing her little sister's antics too well, Iris takes no more than a second to give Alice a correct answer. Can you do the same?

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 50$) — the number of lights in the circuit.

The second line of each test case contains $2n$ integers $a_1, a_2, \ldots, a_{2n}$ ($0 \le a_i \le 1$) — the states of the switches in the circuit. $a_i = 0$ means the $i$-th switch is off, and $a_i = 1$ means the $i$-th switch is on.

Output: For each test case, output two integers — the minimum and maximum number of lights, respectively, that can be turned on.

Examples:
input:
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
output:
0 0
1 1
0 0
0 2
1 3

Note: In the first test case, there is only one light in the circuit, and no switch is on, so the light is certainly off.

In the second test case, there is only one light in the circuit, but one switch connected to it is on, so the light is on.

In the third test case, there is only one light in the circuit, and both switches are on, so the light is off as it was toggled twice.

In the fourth test case, to have no lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $4$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $2$ and switch $6$ are connected to light $2$. Since both switches are off, light $2$ is also off.    * Switch $3$ and switch $5$ are connected to light $3$. Both switches are on, so light $3$ is toggled twice from its initial off state, and thus also stays off.

And to have $2$ lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $2$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $3$ and switch $4$ are connected to light $2$. Since switch $3$ is on and switch $4$ is off, light $2$ is toggled once from its initial off state, so it is on.    * Switch $5$ and switch $6$ are connected to light $3$. Since switch $5$ is on and switch $6$ is off, light $3$ is toggled once from its initial off state, so it is on.

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

using ll = long long;
using vll = vector<ll>;
using vi = vector<int>;
using pii = pair<int, int>;
#define F first
#define S second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()

const int INF = 1e9 + 7;
const int MOD = 1e9 + 7;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t; cin >> t;
    while (t--) {
        int n; cin >> n;
        int mx = 0, mn = 0;
        for (int i = 0; i < 2 * n; i++) {
            int x; cin >> x;
            if (x == 1) {
                mn ^= 1;
                if (mx >= n) mx--;
                else mx++;
            }
        }
        cout << mn << ' ' << mx << '\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:
```plaintext
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Circuit
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice has just crafted a circuit with $n$ lights and $2n$ switches. Each component (a light or a switch) has two states: on or off. The lights and switches are arranged in a way that:

  * Each light is connected to exactly two switches.    * Each switch is connected to exactly one light. It's unknown which light each switch is connected to.    * When all switches are off, all lights are also off.    * If a switch is toggled (from on to off, or vice versa), the state of the light connected to it will also toggle.

Alice brings the circuit, which shows only the states of the $2n$ switches, to her sister Iris and gives her a riddle: what is the minimum and maximum number of lights that can be turned on?

Knowing her little sister's antics too well, Iris takes no more than a second to give Alice a correct answer. Can you do the same?

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 500$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 50$) — the number of lights in the circuit.

The second line of each test case contains $2n$ integers $a_1, a_2, \ldots, a_{2n}$ ($0 \le a_i \le 1$) — the states of the switches in the circuit. $a_i = 0$ means the $i$-th switch is off, and $a_i = 1$ means the $i$-th switch is on.

Output: For each test case, output two integers — the minimum and maximum number of lights, respectively, that can be turned on.

Examples:
input:
5
1
0 0
1
0 1
1
1 1
3
0 0 1 0 1 0
3
0 1 1 1 0 0
output:
0 0
1 1
0 0
0 2
1 3

Note: In the first test case, there is only one light in the circuit, and no switch is on, so the light is certainly off.

In the second test case, there is only one light in the circuit, but one switch connected to it is on, so the light is on.

In the third test case, there is only one light in the circuit, and both switches are on, so the light is off as it was toggled twice.

In the fourth test case, to have no lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $4$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $2$ and switch $6$ are connected to light $2$. Since both switches are off, light $2$ is also off.    * Switch $3$ and switch $5$ are connected to light $3$. Both switches are on, so light $3$ is toggled twice from its initial off state, and thus also stays off.

And to have $2$ lights on, the switches can be arranged in this way:

  * Switch $1$ and switch $2$ are connected to light $1$. Since both switches are off, light $1$ is also off.    * Switch $3$ and switch $4$ are connected to light $2$. Since switch $3$ is on and switch $4$ is off, light $2$ is toggled once from its initial off state, so it is on.    * Switch $5$ and switch $6$ are connected to light $3$. Since switch $5$ is on and switch $6$ is off, light $3$ is toggled once from its initial off state, so it is on.

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

int32_t main(){
	int tc;
	cin>>tc;
	while(tc--){
		int n;
		cin>>n;
		vector<int>a(2*n);
		int val = 0;
		for(int i=0;i<2*n;i++) {
			cin>>a[i];
			if(a[i]==1) val++;
		}
		if(n==1){
			if(val==1) cout<<1<<" "<<1<<endl;
			else cout<<0<<" "<<0<<endl;
			continue;
		}
		int maxi = min(val,n);
		if(val>n) maxi -= val%n;
		if(val==2*n) maxi -= 2*n;
		int mini = val%2;
		cout<<mini<<" "<<maxi<<endl;
	}
	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 buggy code, we need to focus on edge cases and scenarios that highlight the incorrect logic in the code. The buggy code has issues with calculating the maximum and minimum number of lights that can be turned on. Here are some key points to consider:

1. **Case with all switches off**: The minimum and maximum number of lights should be 0.
2. **Case with all switches on**: The minimum number of lights should be 0 (since each light is toggled twice), and the maximum number of lights should be 0 as well.
3. **Case with an odd number of switches on**: The minimum number of lights should be 1 (since at least one light will be toggled once), and the maximum number of lights should be the number of lights that can be toggled once.
4. **Case with an even number of switches on but not all**: The minimum number of lights should be 0 (since the toggles can be paired to cancel out), and the maximum number of lights should be the number of lights that can be toggled once.

Given these points, let's create a small-scale test input:

```plaintext
4
1
0 0
1
1 1
2
1 1 0 0
2
1 0 1 0
```

### Explanation:
1. **Test Case 1**:
   - \( n = 1 \)
   - Switches: 0 0
   - Expected output: 0 0
   - This tests the case where all switches are off.

2. **Test Case 2**:
   - \( n = 1 \)
   - Switches: 1 1
   - Expected output: 0 0
   - This tests the case where all switches are on.

3. **Test Case 3**:
   - \( n = 2 \)
   - Switches: 1 1 0 0
   - Expected output: 0 2
   - This tests the case where an even number of switches are on but not all.

4. **Test Case 4**:
   - \( n = 2 \)
   - Switches: 1 0 1 0
   - Expected output: 1 2
   - This tests the case where an odd number of switches are on.

These test cases should help expose the bugs in the provided code.