**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Action Figures
time_limit_ms: 2500
memory_limit_mb: 512
Description: There is a shop that sells action figures near Monocarp's house. A new set of action figures will be released shortly; this set contains $n$ figures, the $i$-th figure costs $i$ coins and is available for purchase from day $i$ to day $n$.

For each of the $n$ days, Monocarp knows whether he can visit the shop.

Every time Monocarp visits the shop, he can buy any number of action figures which are sold in the shop (of course, he cannot buy an action figure that is not yet available for purchase). If Monocarp buys at least two figures during the same day, he gets a discount equal to the cost of the most expensive figure he buys (in other words, he gets the most expensive of the figures he buys for free).

Monocarp wants to buy exactly one $1$-st figure, one $2$-nd figure, ..., one $n$-th figure from the set. He cannot buy the same figure twice. What is the minimum amount of money he has to spend?

Input: The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($1 \le n \le 4 \cdot 10^5$) — the number of figures in the set (and the number of days);    * the second line contains a string $s$ ($|s| = n$, each $s_i$ is either 0 or 1). If Monocarp can visit the shop on the $i$-th day, then $s_i$ is 1; otherwise, $s_i$ is 0.

Additional constraints on the input:

  * in each test case, $s_n$ is 1, so Monocarp is always able to buy all figures during the $n$-th day;    * the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, print one integer — the minimum amount of money Monocarp has to spend.

Examples:
input:
4
1
1
6
101101
7
1110001
5
11111
output:
1
8
18
6

Note: In the first test case, Monocarp buys the $1$-st figure on the $1$-st day and spends $1$ coin.

In the second test case, Monocarp can buy the $1$-st and the $3$-rd figure on the $3$-rd day, the $2$-nd and the $4$-th figure on the $4$-th day, and the $5$-th and the $6$-th figure on the $6$-th day. Then, he will spend $1+2+5=8$ coins.

In the third test case, Monocarp can buy the $2$-nd and the $3$-rd figure on the $3$-rd day, and all other figures on the $7$-th day. Then, he will spend $1+2+4+5+6 = 18$ coins.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int MOD = 1e9;
#define intt int
#define pll pair<long long, long long>
#define vpll vector<pair<long long, long long>>
#define vll vector<long long>
#define mll map<ll, ll>
#define si set<int>
#define sc set<char>
#define ff first
#define se second
#define all(v) (v).begin(), (v).end()
#define allr(v) (v).rbegin(), (v).rend()
#define f(i, s, e) for (long long int i = s; i < e; i++)
#define cf(i, s, e) for (long long int i = s; i <= e; i++)
#define rf(i, e, s) for (long long int i = e - 1; i >= s; i--)
#define pb push_back
#define eb emplace_back
#define M 1000000007
#define PI 3.1415926535897932384626433832795
#define read(type) readInt<type>()
#define take(n, a) vll a(n); f(i, 0, n) cin >> a[i];
ll min(ll a, ll b) { if (a < b) return a; return b; }
ll max(ll a, ll b) { if (a > b) return a; return b; }
ll gcd(ll a, ll b) { if (b > a) return gcd(b, a); if (b == 0) return a; return gcd(b, a % b); }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
string to_upper(string a) { for (int i = 0; i < (int)a.size(); ++i) if (a[i] >= 'a' && a[i] <= 'z') a[i] -= 'a' - 'A'; return a; }
string to_lower(string a) { for (int i = 0; i < (int)a.size(); ++i) if (a[i] >= 'A' && a[i] <= 'Z') a[i] += 'a' - 'A'; return a; }
bool prime(int a) { if (a == 1) return 0; for (int i = 2; i <= round(sqrt(a)); ++i) if (a % i == 0) return 0; return 1; }
void yes() { cout << "YES\n"; }
void no() { cout << "NO\n"; }

class DisjointSet {
    vector<int> rank, parent, size;
public:
    DisjointSet(int n) {
        rank.resize(n + 1, 0);
        parent.resize(n + 1);
        size.resize(n + 1);
        for (int i = 0; i <= n; i++) {
            parent[i] = i;
            size[i] = 1;
        }
    }

    int findUPar(int node) {
        if (node == parent[node])
            return node;
        return parent[node] = findUPar(parent[node]);
    }

    void unionBySize(int u, int v) {
        int ulp_u = findUPar(u);
        int ulp_v = findUPar(v);
        if (ulp_u == ulp_v) return;
        if (size[ulp_u] < size[ulp_v]) {
            parent[ulp_u] = ulp_v;
            size[ulp_v] += size[ulp_u];
        }
        else {
            parent[ulp_v] = ulp_u;
            size[ulp_u] += size[ulp_v];
        }
    }
};
long long binpow(long long a, long long b) {
    long long res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a;
        a = a * a;
        b >>= 1;
    }
    return res;
}

int main() {
   ll t;
   cin>>t;
   while(t--)
   {
    ll n;
    cin>>n;
    string s;
    cin>>s;
    int fl=0;
    ll ans=0;
    vector<int> v;
    for(int i=0;i<n;i++)
    {
        if(s[i]=='0')
        {
            fl=1;
            ans+=i+1;
        }
        else
        {
            if(fl==0)
            v.push_back(i+1);
            fl=0;

        }
    }
    int k=v.size()/2;
    if(v.size())
    {
    for(int i=0;i<=(v.size()/2);i++)
    ans+=v[i];
    if(v.size()%2==0)
    ans-=v[v.size()/2];
    }
    cout<<ans<<endl;
   }
}

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

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

Think step by step.

Response:
```plaintext
1
5
00001
```

### Explanation:
This test case is designed to expose potential issues in the logic of the buggy code.

1. **Input Construction**:
   - The number of test cases is 1.
   - The number of figures (and days) is 5.
   - The string `s` is "00001", indicating that Monocarp can only visit the shop on the 5th day.

2. **Expected Behavior**:
   - Since Monocarp can only visit the shop on the 5th day, he must buy all the figures on that day.
   - The cost of buying all figures on the 5th day is the sum of the costs of the figures minus the cost of the most expensive figure (since he gets the most expensive figure for free).

3. **Correct Calculation**:
   - The total cost of all figures: \(1 + 2 + 3 + 4 + 5 = 15\).
   - The most expensive figure is 5.
   - The actual cost Monocarp pays is \(15 - 5 = 10\).

4. **Buggy Code Analysis**:
   - The buggy code accumulates the cost of all figures that are not available on their respective days and then adds the cost of some figures from the available days.
   - The logic in the buggy code might incorrectly handle the case where all figures are bought on the last day, leading to an incorrect total cost.

5. **Expected Output**:
   - The correct output for this test case should be 10.

By using this test case, we can verify if the buggy code correctly handles the scenario where Monocarp can only visit the shop on the last day, and thus must buy all figures on that day with the appropriate discount.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Action Figures
time_limit_ms: 2500
memory_limit_mb: 512
Description: There is a shop that sells action figures near Monocarp's house. A new set of action figures will be released shortly; this set contains $n$ figures, the $i$-th figure costs $i$ coins and is available for purchase from day $i$ to day $n$.

For each of the $n$ days, Monocarp knows whether he can visit the shop.

Every time Monocarp visits the shop, he can buy any number of action figures which are sold in the shop (of course, he cannot buy an action figure that is not yet available for purchase). If Monocarp buys at least two figures during the same day, he gets a discount equal to the cost of the most expensive figure he buys (in other words, he gets the most expensive of the figures he buys for free).

Monocarp wants to buy exactly one $1$-st figure, one $2$-nd figure, ..., one $n$-th figure from the set. He cannot buy the same figure twice. What is the minimum amount of money he has to spend?

Input: The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($1 \le n \le 4 \cdot 10^5$) — the number of figures in the set (and the number of days);    * the second line contains a string $s$ ($|s| = n$, each $s_i$ is either 0 or 1). If Monocarp can visit the shop on the $i$-th day, then $s_i$ is 1; otherwise, $s_i$ is 0.

Additional constraints on the input:

  * in each test case, $s_n$ is 1, so Monocarp is always able to buy all figures during the $n$-th day;    * the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, print one integer — the minimum amount of money Monocarp has to spend.

Examples:
input:
4
1
1
6
101101
7
1110001
5
11111
output:
1
8
18
6

Note: In the first test case, Monocarp buys the $1$-st figure on the $1$-st day and spends $1$ coin.

In the second test case, Monocarp can buy the $1$-st and the $3$-rd figure on the $3$-rd day, the $2$-nd and the $4$-th figure on the $4$-th day, and the $5$-th and the $6$-th figure on the $6$-th day. Then, he will spend $1+2+5=8$ coins.

In the third test case, Monocarp can buy the $2$-nd and the $3$-rd figure on the $3$-rd day, and all other figures on the $7$-th day. Then, he will spend $1+2+4+5+6 = 18$ coins.

Buggy Code:
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <typename T> using pbds = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// order_of_key(k) : no. of elements < k
// *find_by_order(i) : value at index i (0-based)
#define ll long long
#define fast cin.tie(0), cout.tie(0), cin.sync_with_stdio(0), cout.sync_with_stdio(0);
#define endl "\n"
#define in(n) ll n; cin >> n
#define out(n) cout << n << endl
#define invec(vec,n) for(int i=0;i<n;i++){int p;cin >> p;vec.push_back(p);}
#define fr(a,b) for(ll i=a;i<b;i++)
#define rfr(a,b) for(ll i=a;i>b;i--)
#define srt(v) sort(v.begin(),v.end())
#define asor(ar) sort(ar,ar+ar.size())
#define rsrt(vec) sort(vec.rbegin(),vec.rend())
#define vll vector<ll>
#define pb push_back
#define pob pop_back
#define eb emplace_back
#define setbit(x) __builtin_popcountll(x)
#define mostsigbit(x) __builtin_clzll(x)
#define parity(x) __builtin_parityll(x) // returns true(1) if no has odd no's of 1 in binary else return false(0)
#define firstsetbit(x) __builtin_ctzll(x)
#define M_PI 3.14159265358979323846
#define all(x) x.begin(),x.end()
#define nxt_per(v) next_permutation(v.begin(),v.end())
#define lcm(a,b) (a/__gcd(a, b)*b)
// STL DECLARATION SHORTCUTS
#define pii pair<int,int>
#define vi vector<int>
#define vc vector<char>
#define vs vector<string>
#define vvi vector<vector<int>>
#define vvc vector<vector<char>>
#define vpii vector<pair<int,int>>
#define vvpii vector<vector<pair<int,int>>>
#define mii map<int,int>
#define mci map<char,int>
#define msi map<string,int>
#define mivi map<int,vector<int>>
#define si set<int>
#define sc set<char>
#define ss set<string>
#define qi queue<int>
#define dqi deque<int>
#define stki stack<int>

// MOST COMMON OUTPUTS
#define YES cout << "YES" << endl
#define NO cout << "NO" << endl
#define Yes cout << "Yes" << endl
#define No cout << "No" << endl
#define outarr(arr) for(int i=0;i<(sizeof(arr)/sizeof(arr[0]));i++){ cout << arr[i] << " ";cout << endl;}
#define outvec(v) for(int i=0;i<v.size();i++){ cout << v[i] << " ";}cout << endl;
#define outmap(m) for(auto i: m){ cout << i.first << " " << i.second << endl;}
#define outvecvec(v) for(int i=0;i<v.size();i++){for(int j=0;j<v[i].size();j++){cout << v[i][j] << " ";}cout << endl;}
#define outvecpair(v) for(int i=0;i<v.size();i++){cout << v[i].first << " " << v[i].second << endl;}

const ll MOD = 998244353;
#define mod 1000000007
#define int long long
int ceildiv(int a, int b) {return (a + b - 1) / b;}
void solve(void)
{
   in(n);
   string s;
   cin>>s;
   int ans=0;
   for(int i=0;i<n;i++)
   {
      if(s[i]=='0')
      ans+=i+1;
   }
   vi v;
   for(int i=0;i<n;i++)
   {
      if((i==0 and s[i]=='1') or  (s[i]=='1' and s[i-1]!='0'))
      v.pb(i+1);
   }
   int sz=v.size();
   for(int i=0;i<(sz+1)/2;i++)
   ans+=v[i];
   out(ans);
}
int32_t main()
{
fast
int t=1;
cin>>t;
while(t--)
{
solve();
}
}
//Avoid using pow function for even moderately larger values as it gives precision errors , BETTER TO USE WHILE LOOP TO GET POWER OF ANY NUMBER.
//Whenever you write (O(N^2)) code, then there always exists an (O(N)) code or (O(Nlog N)) code that can be optimized using two pointers or binary search or STL, or a mixture of any of these two.
//For range queries, always think of precomputation (like prefix sum) before going into the queries.
// for interactive problems--> 1. your output is input to interact and systems output after interaction is your input
// 2. any time you write cout<<... always use endl other wise idleness limit will exceed 100% and make sure to not have endl defined as '/n'
// 3. no compiler will give you the correct final output so rather check output of steps and good luck for submission

**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
4
1
1
6
101101
7
1110001
5
11111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Action Figures
time_limit_ms: 2500
memory_limit_mb: 512
Description: There is a shop that sells action figures near Monocarp's house. A new set of action figures will be released shortly; this set contains $n$ figures, the $i$-th figure costs $i$ coins and is available for purchase from day $i$ to day $n$.

For each of the $n$ days, Monocarp knows whether he can visit the shop.

Every time Monocarp visits the shop, he can buy any number of action figures which are sold in the shop (of course, he cannot buy an action figure that is not yet available for purchase). If Monocarp buys at least two figures during the same day, he gets a discount equal to the cost of the most expensive figure he buys (in other words, he gets the most expensive of the figures he buys for free).

Monocarp wants to buy exactly one $1$-st figure, one $2$-nd figure, ..., one $n$-th figure from the set. He cannot buy the same figure twice. What is the minimum amount of money he has to spend?

Input: The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($1 \le n \le 4 \cdot 10^5$) — the number of figures in the set (and the number of days);    * the second line contains a string $s$ ($|s| = n$, each $s_i$ is either 0 or 1). If Monocarp can visit the shop on the $i$-th day, then $s_i$ is 1; otherwise, $s_i$ is 0.

Additional constraints on the input:

  * in each test case, $s_n$ is 1, so Monocarp is always able to buy all figures during the $n$-th day;    * the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, print one integer — the minimum amount of money Monocarp has to spend.

Examples:
input:
4
1
1
6
101101
7
1110001
5
11111
output:
1
8
18
6

Note: In the first test case, Monocarp buys the $1$-st figure on the $1$-st day and spends $1$ coin.

In the second test case, Monocarp can buy the $1$-st and the $3$-rd figure on the $3$-rd day, the $2$-nd and the $4$-th figure on the $4$-th day, and the $5$-th and the $6$-th figure on the $6$-th day. Then, he will spend $1+2+5=8$ coins.

In the third test case, Monocarp can buy the $2$-nd and the $3$-rd figure on the $3$-rd day, and all other figures on the $7$-th day. Then, he will spend $1+2+4+5+6 = 18$ coins.

Buggy Code:
def run_case():
    n = int(input())
    s = "0" + input().strip()
    left, right = 1, n
    cost = 0
    while right > 0:
        if left > right:
            break
        elif left == right:
            cost += right
            break
        middle = right - 1
        while middle > left and s[middle] == "0":
            cost += middle
            middle -= 1
        if middle == right - 1:
            cost += left
            left += 1
        right = middle
    print(cost)

    return

t = int(input())
for _ in range(t):
    run_case()

**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
4
1
1
2
10
3
101
4
1111
```

### Explanation:
1. **Test Case 1:**
   - **Input:** `1 1`
   - **Output:** `1`
   - **Explanation:** Monocarp buys the 1st figure on the 1st day, spending 1 coin.

2. **Test Case 2:**
   - **Input:** `2 10`
   - **Output:** `2`
   - **Explanation:** Monocarp can only buy the 2nd figure on the 2nd day, spending 2 coins. He cannot buy the 1st figure because he cannot visit the shop on the 1st day.

3. **Test Case 3:**
   - **Input:** `3 101`
   - **Output:** `4`
   - **Explanation:** Monocarp buys the 1st and 3rd figures on the 3rd day, spending 1 + 2 = 3 coins (the 3rd figure is free due to the discount). He cannot buy the 2nd figure on the 2nd day because he cannot visit the shop on the 2nd day.

4. **Test Case 4:**
   - **Input:** `4 1111`
   - **Output:** `6`
   - **Explanation:** Monocarp can buy the 1st and 2nd figures on the 2nd day, spending 1 coin (the 2nd figure is free due to the discount). Then, he buys the 3rd and 4th figures on the 4th day, spending 3 coins (the 4th figure is free due to the discount). Total cost is 1 + 3 = 6 coins.

These test cases are designed to cover edge cases and scenarios where the buggy code might fail, such as when Monocarp can only visit the shop on the last day, or when he can visit the shop on multiple days but the sequence of visits is important.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Action Figures
time_limit_ms: 2500
memory_limit_mb: 512
Description: There is a shop that sells action figures near Monocarp's house. A new set of action figures will be released shortly; this set contains $n$ figures, the $i$-th figure costs $i$ coins and is available for purchase from day $i$ to day $n$.

For each of the $n$ days, Monocarp knows whether he can visit the shop.

Every time Monocarp visits the shop, he can buy any number of action figures which are sold in the shop (of course, he cannot buy an action figure that is not yet available for purchase). If Monocarp buys at least two figures during the same day, he gets a discount equal to the cost of the most expensive figure he buys (in other words, he gets the most expensive of the figures he buys for free).

Monocarp wants to buy exactly one $1$-st figure, one $2$-nd figure, ..., one $n$-th figure from the set. He cannot buy the same figure twice. What is the minimum amount of money he has to spend?

Input: The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($1 \le n \le 4 \cdot 10^5$) — the number of figures in the set (and the number of days);    * the second line contains a string $s$ ($|s| = n$, each $s_i$ is either 0 or 1). If Monocarp can visit the shop on the $i$-th day, then $s_i$ is 1; otherwise, $s_i$ is 0.

Additional constraints on the input:

  * in each test case, $s_n$ is 1, so Monocarp is always able to buy all figures during the $n$-th day;    * the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, print one integer — the minimum amount of money Monocarp has to spend.

Examples:
input:
4
1
1
6
101101
7
1110001
5
11111
output:
1
8
18
6

Note: In the first test case, Monocarp buys the $1$-st figure on the $1$-st day and spends $1$ coin.

In the second test case, Monocarp can buy the $1$-st and the $3$-rd figure on the $3$-rd day, the $2$-nd and the $4$-th figure on the $4$-th day, and the $5$-th and the $6$-th figure on the $6$-th day. Then, he will spend $1+2+5=8$ coins.

In the third test case, Monocarp can buy the $2$-nd and the $3$-rd figure on the $3$-rd day, and all other figures on the $7$-th day. Then, he will spend $1+2+4+5+6 = 18$ coins.

Buggy Code:
def ii():
    return int(input())
def li():
    return list(map(int,input().split()))
def si():
    return input()
def ti():
    return map(int,input().split())
def di():
    return deque((map(int,input().split())))
from math import *
from bisect import bisect_left,bisect_right
from heapq import heapify,heappop,heappush
from itertools import permutations
from collections import deque,defaultdict,Counter
def maxsubarraysum(arr,k):
    sumi=0
    best=0
    for i in range(k):
        sumi=max(arr[i],arr[i]+sumi)
        best=max(best,sumi)
    return best
def mod_inverse(a, m):
    m0 = m
    y = 0
    x = 1
    if m == 1:
        return 0
    while a > 1:
        q = a // m
        t = m
        m = a % m
        a = t
        t = y
        y = x - q * y
        x = t
    if x < 0:
        x = x + m0
    return x
def prefixsum(l,n):
    if n==0:
        return []
    array=[l[0]]
    for i in range(1,n):
        array.append(array[-1]+l[i])
    return array
def suffixsum(l,n):
    array=[0]*n
    temperory=0
    for i in range(n):
        temperory+=l[n-1-i]
        array[n-1-i]=temperory
    return array
def prime(x):
    if x==1:
        return(0)
    elif x==2:
        return(1)
    else:
        t=(sqrt(x))+1
        t=int(t)
        for i in range(2,t):
            if x%i==0:
                return(0)
        else:
            return(1)
def prefixmax(l,n):
    array=[l[0]]
    for i in range(1,n):
        array.append(max(array[-1],l[i]))
    return array
def prefixmin(l,n):
    array=[l[0]]
    for i in range(1,n):
        array.append(min(array[-1],l[i]))
    return array
def BinarySearch(a, x):

    i = bisect_left(a, x)

    if i != len(a) and a[i] == x:

        return i

    else:

        return -1
def sortremoveDuplicates(arr, n):
    if n == 0 or n == 1:
        return arr

    temp = []
    j = 0
    for i in range(0, n-1):
        if arr[i] != arr[i+1]:
            temp += [arr[i]]
            j += 1
    temp += [arr[n-1]]
    return temp
def decimalToBinary(n):
    return bin(n).replace("0b", "")
def ato1(chr):
    return ord(chr)-ord("A")
def permutationindex(l,n):
    arr=[0]*(n)
    for i in range(n):
        arr[l[i]-1]=i
    return arr
def issubsequence(s, s2):

    n, m = len(s), len(s2)
    i, j = 0, 0
    while (i < n and j < m):
        if (s[i] == s2[j]):
            i += 1
        j += 1

    # If i reaches end of s,that mean we found all
    # characters of s in s2,
    # so s is subsequence of s2, else not
    return i == n
def def_value():
    return -1
def primef(n):
    l1=[]
    i=2
    while i*i<=n:
        if n%i!=0:
            i+=1
        else:
            l1.append(i)
            while(n%i==0):
                n//=i
    if n>1:
        l1.append(n)
    return l1
"""
def search( k) :
        global ans
        if (k == len(l)) :
           if len(subset)%2==0 and len(subset)>0:
               ans+=maxi//product(subset)
           elif len(subset)>0:
               ans-=maxi//product(subset)
        else:
           search(k+1)
           subset.append(l[k])
           search(k+1)
           subset.pop()
"""
def seive(n):
    ans=[0]*(n+1)
    #primes=[]
    for i in range(2,n+1):
        if ans[i]==0:
            #primes.append(i)
            for j in range(2*i,n+1,i):
                ans[j]=i
    return ans
def inte(s,s2):
    if s[1]<s2[0] or s2[1]<s[0]:
        return 0
    return [max(s[0],s2[0]),min(s[1],s2[1])]
def checksq(x):
    z=isqrt(x)
    if z<0:
        return False
    if z*z==x:
        return True
    return False
def counti(l,n):
    l1=[]
    z=1
    for i in range(1,n):
        if l[i]==l[i-1]:
            z+=1
        else:
            l1.append([l[i-1],z])
            z=1
    l1.append([l[-1],z])
    return l1
def next(arr, target):
    start = 0;
    end = len(arr) - 1;
    # Minimum size of the array should be 1
    if (end == 0):
        return -1
    '''
    If target lies beyond the max element, than the index of strictly smaller
    value than target should be (end - 1)
    '''
    if (target > arr[end]):
        return end

    ans = -1;
    while (start <= end):
        mid = (start + end) // 2;
        if (arr[mid] >= target):
            end = mid - 1;
        else:
            ans = mid;
            start = mid + 1;

    return ans;
def next1(arr, target):
    start = 0
    end = len(arr) - 1

    ans = -1
    while (start <= end):
        mid = (start + end) // 2

        # Move to right side if target is
        # greater.
        if (arr[mid] <= target):
            start = mid + 1

        # Move left side.
        else:
            ans = mid
            end = mid - 1

    return ans
def next_greatest_indices(arr):
    result = [len(arr)] * len(arr)
    stack = []
    for i in range(len(arr) - 1, -1, -1):
        while stack and arr[stack[-1]] < arr[i]:
            stack.pop()
        if stack:
            result[i] = stack[-1]
        stack.append(i)
    return result
def find_largest_index(arr, k):
    left, right = 0, len(arr) - 1
    result = -1

    while left <= right:
        mid = (left + right) // 2

        if arr[mid] < k:
            result = mid
            left = mid + 1
        else:
            right = mid - 1

    return result

import sys
input=sys.stdin.readline
for i in range(ii()):
    n=ii()
    s=si()
    ans=n*(n+1)
    ans//=2
    l=[1]
    j=0
    i=1
    while (i<n):
        z=i
        if (s[i]=='0'):
            l.append(i+1)
            i+=1
            continue
        while (z<n and s[z]=='1'):
            z+=1
        zz=z-1
        while (zz>=i):
            if (j<len(l)):
                ans-=zz+1
                j+=1
                zz-=1
            elif (zz-1>=i):
                ans-=zz+1
                zz-=2
            else:
                l.append(i)
                zz-=1
        i=z
    print(ans)

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

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

Think step by step.

Response:
```plaintext
4
1
1
6
101101
7
1110001
5
11111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Action Figures
time_limit_ms: 2500
memory_limit_mb: 512
Description: There is a shop that sells action figures near Monocarp's house. A new set of action figures will be released shortly; this set contains $n$ figures, the $i$-th figure costs $i$ coins and is available for purchase from day $i$ to day $n$.

For each of the $n$ days, Monocarp knows whether he can visit the shop.

Every time Monocarp visits the shop, he can buy any number of action figures which are sold in the shop (of course, he cannot buy an action figure that is not yet available for purchase). If Monocarp buys at least two figures during the same day, he gets a discount equal to the cost of the most expensive figure he buys (in other words, he gets the most expensive of the figures he buys for free).

Monocarp wants to buy exactly one $1$-st figure, one $2$-nd figure, ..., one $n$-th figure from the set. He cannot buy the same figure twice. What is the minimum amount of money he has to spend?

Input: The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($1 \le n \le 4 \cdot 10^5$) — the number of figures in the set (and the number of days);    * the second line contains a string $s$ ($|s| = n$, each $s_i$ is either 0 or 1). If Monocarp can visit the shop on the $i$-th day, then $s_i$ is 1; otherwise, $s_i$ is 0.

Additional constraints on the input:

  * in each test case, $s_n$ is 1, so Monocarp is always able to buy all figures during the $n$-th day;    * the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, print one integer — the minimum amount of money Monocarp has to spend.

Examples:
input:
4
1
1
6
101101
7
1110001
5
11111
output:
1
8
18
6

Note: In the first test case, Monocarp buys the $1$-st figure on the $1$-st day and spends $1$ coin.

In the second test case, Monocarp can buy the $1$-st and the $3$-rd figure on the $3$-rd day, the $2$-nd and the $4$-th figure on the $4$-th day, and the $5$-th and the $6$-th figure on the $6$-th day. Then, he will spend $1+2+5=8$ coins.

In the third test case, Monocarp can buy the $2$-nd and the $3$-rd figure on the $3$-rd day, and all other figures on the $7$-th day. Then, he will spend $1+2+4+5+6 = 18$ coins.

Buggy Code:
for _ in range(0,int(input())):
    n=int(input())
    s=[bool(int(l)) for l in input()]

    ct_0=0
    position_0=[-1]*(n+1)
    for i in range(0,n):
        if not(s[i]):
            ct_0+=1
            position_0[ct_0]=i

    discount_money=0
    for i in range(n-1,-1,-1):
        if not(s[i]): continue
        discnt=False
        while ct_0>0:
            if position_0[ct_0]<i:
                discnt=True
                ct_0-=1
                break
            else:
                ct_0-=1
        if discnt:
            discount_money+=(i+1)
            continue
        else:
            ed=i
            break
    st=0
    while ed>st:
        while ed>=0 and s[ed]==0:
            ed-=1
        while st<n-1 and s[st]==0:
            st+=1
        if ed>st:
            discount_money+=(ed+1)
        ed-=1
        st+=1
    print(n*(n+1)//2-discount_money)

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

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

Think step by step.

Response:
```plaintext
3
1
1
2
11
3
101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Action Figures
time_limit_ms: 2500
memory_limit_mb: 512
Description: There is a shop that sells action figures near Monocarp's house. A new set of action figures will be released shortly; this set contains $n$ figures, the $i$-th figure costs $i$ coins and is available for purchase from day $i$ to day $n$.

For each of the $n$ days, Monocarp knows whether he can visit the shop.

Every time Monocarp visits the shop, he can buy any number of action figures which are sold in the shop (of course, he cannot buy an action figure that is not yet available for purchase). If Monocarp buys at least two figures during the same day, he gets a discount equal to the cost of the most expensive figure he buys (in other words, he gets the most expensive of the figures he buys for free).

Monocarp wants to buy exactly one $1$-st figure, one $2$-nd figure, ..., one $n$-th figure from the set. He cannot buy the same figure twice. What is the minimum amount of money he has to spend?

Input: The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($1 \le n \le 4 \cdot 10^5$) — the number of figures in the set (and the number of days);    * the second line contains a string $s$ ($|s| = n$, each $s_i$ is either 0 or 1). If Monocarp can visit the shop on the $i$-th day, then $s_i$ is 1; otherwise, $s_i$ is 0.

Additional constraints on the input:

  * in each test case, $s_n$ is 1, so Monocarp is always able to buy all figures during the $n$-th day;    * the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, print one integer — the minimum amount of money Monocarp has to spend.

Examples:
input:
4
1
1
6
101101
7
1110001
5
11111
output:
1
8
18
6

Note: In the first test case, Monocarp buys the $1$-st figure on the $1$-st day and spends $1$ coin.

In the second test case, Monocarp can buy the $1$-st and the $3$-rd figure on the $3$-rd day, the $2$-nd and the $4$-th figure on the $4$-th day, and the $5$-th and the $6$-th figure on the $6$-th day. Then, he will spend $1+2+5=8$ coins.

In the third test case, Monocarp can buy the $2$-nd and the $3$-rd figure on the $3$-rd day, and all other figures on the $7$-th day. Then, he will spend $1+2+4+5+6 = 18$ coins.

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

import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.System.*;

public class Main {
    public static Scanner sc = new Scanner(in);
    public static PrintWriter pw = new PrintWriter(out);

    public static void main(String[] args) throws IOException {

        int test = 1;
        test = sc.nextInt();
        while (test-- > 0) {
            int n = sc.nextInt();
            String s = sc.next();
            int cnt=0;
            int cnt2=0;
            int out=0;
            int[] a=new int[n];
            for(int i=0;i<n-1;i++){
                if(s.charAt(i)=='0'){
                    cnt++;
                    out+=i+1;
                }
                else if(s.charAt(i)=='1' && s.charAt(i+1)=='0' && cnt>0){
                    cnt--;
                }
                else{
                    a[i]=i+1;
                    cnt2++;
                }
            }
            if(s.charAt(n-1)=='1' && cnt<=0){
                cnt2++;
                a[n-1]=n;
            }

            cnt2=(cnt2+1)/2;
            for(int i=0;i<n && cnt2>0;i++){
                if(a[i]>0)
                    cnt2--;
                out+=a[i];

            }
            pw.println(out);
        }
        pw.flush();

    }
    static class Scanner {

        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(String file) throws IOException {
            br = new BufferedReader(new FileReader(file));
        }

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

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

        public String readAllLines(BufferedReader reader) throws IOException {
            StringBuilder content = new StringBuilder();
            String line;
            while ((line = reader.readLine()) != null) {
                content.append(line);
                content.append(lineSeparator());
            }
            return content.toString();
        }

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec) f *= 10;
                }
            res += parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public long[] nextlongarr(int n) throws IOException {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public Long[] nextLongarr(int n) throws IOException {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public int[] nextIntarr(int n) throws IOException {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public Integer[] nextIntegerarr(int n) throws IOException {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public boolean ready() throws IOException {
            return br.ready();
        }

    }

}

**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
4
1
1
6
101101
7
1110001
5
11111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Action Figures
time_limit_ms: 2500
memory_limit_mb: 512
Description: There is a shop that sells action figures near Monocarp's house. A new set of action figures will be released shortly; this set contains $n$ figures, the $i$-th figure costs $i$ coins and is available for purchase from day $i$ to day $n$.

For each of the $n$ days, Monocarp knows whether he can visit the shop.

Every time Monocarp visits the shop, he can buy any number of action figures which are sold in the shop (of course, he cannot buy an action figure that is not yet available for purchase). If Monocarp buys at least two figures during the same day, he gets a discount equal to the cost of the most expensive figure he buys (in other words, he gets the most expensive of the figures he buys for free).

Monocarp wants to buy exactly one $1$-st figure, one $2$-nd figure, ..., one $n$-th figure from the set. He cannot buy the same figure twice. What is the minimum amount of money he has to spend?

Input: The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($1 \le n \le 4 \cdot 10^5$) — the number of figures in the set (and the number of days);    * the second line contains a string $s$ ($|s| = n$, each $s_i$ is either 0 or 1). If Monocarp can visit the shop on the $i$-th day, then $s_i$ is 1; otherwise, $s_i$ is 0.

Additional constraints on the input:

  * in each test case, $s_n$ is 1, so Monocarp is always able to buy all figures during the $n$-th day;    * the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, print one integer — the minimum amount of money Monocarp has to spend.

Examples:
input:
4
1
1
6
101101
7
1110001
5
11111
output:
1
8
18
6

Note: In the first test case, Monocarp buys the $1$-st figure on the $1$-st day and spends $1$ coin.

In the second test case, Monocarp can buy the $1$-st and the $3$-rd figure on the $3$-rd day, the $2$-nd and the $4$-th figure on the $4$-th day, and the $5$-th and the $6$-th figure on the $6$-th day. Then, he will spend $1+2+5=8$ coins.

In the third test case, Monocarp can buy the $2$-nd and the $3$-rd figure on the $3$-rd day, and all other figures on the $7$-th day. Then, he will spend $1+2+4+5+6 = 18$ coins.

Buggy Code:
import java.util.*;
public class CP {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int tc = sc.nextInt();
        while (tc-- > 0) {
            int n = sc.nextInt();
            String s  = sc.next();
            if(n == 1) {
                System.out.println(s);
                continue;
            }
            int i = n-1;
            int wst = 0;
            int j = n-1;
            while(i>=0){

                while((j>=0 && s.charAt(j) != '0') || j > i){
                    j--;
                }
                if(j>=0){
                    wst+= i+1;
                }
                else{
                    break;
                }
                i--;
                if(i<0){break;}
                 while(i>=0 && s.charAt(i) != '1'){i--;}
                j--;
            }

            if(i>=0 && j<0){
               int c = i/2;

               while(i>=0 && c>0){
                if(s.charAt(i) == '0'){i--;continue;}
                  wst+= i+1;
                  i--;
                  c--;
               }
            }
            int ans = (n*(n+1)/2)-wst;
            System.out.println(ans);
        }
        sc.close();
    }
}

// 4
// 1
// 1
// 6
// 101101
// 7
// 1110001
// 5
// 11111
// OutputCopy
// 1
// 8
// 18
// 6

**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
4
1
1
6
101101
7
1110001
5
11111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Action Figures
time_limit_ms: 2500
memory_limit_mb: 512
Description: There is a shop that sells action figures near Monocarp's house. A new set of action figures will be released shortly; this set contains $n$ figures, the $i$-th figure costs $i$ coins and is available for purchase from day $i$ to day $n$.

For each of the $n$ days, Monocarp knows whether he can visit the shop.

Every time Monocarp visits the shop, he can buy any number of action figures which are sold in the shop (of course, he cannot buy an action figure that is not yet available for purchase). If Monocarp buys at least two figures during the same day, he gets a discount equal to the cost of the most expensive figure he buys (in other words, he gets the most expensive of the figures he buys for free).

Monocarp wants to buy exactly one $1$-st figure, one $2$-nd figure, ..., one $n$-th figure from the set. He cannot buy the same figure twice. What is the minimum amount of money he has to spend?

Input: The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($1 \le n \le 4 \cdot 10^5$) — the number of figures in the set (and the number of days);    * the second line contains a string $s$ ($|s| = n$, each $s_i$ is either 0 or 1). If Monocarp can visit the shop on the $i$-th day, then $s_i$ is 1; otherwise, $s_i$ is 0.

Additional constraints on the input:

  * in each test case, $s_n$ is 1, so Monocarp is always able to buy all figures during the $n$-th day;    * the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, print one integer — the minimum amount of money Monocarp has to spend.

Examples:
input:
4
1
1
6
101101
7
1110001
5
11111
output:
1
8
18
6

Note: In the first test case, Monocarp buys the $1$-st figure on the $1$-st day and spends $1$ coin.

In the second test case, Monocarp can buy the $1$-st and the $3$-rd figure on the $3$-rd day, the $2$-nd and the $4$-th figure on the $4$-th day, and the $5$-th and the $6$-th figure on the $6$-th day. Then, he will spend $1+2+5=8$ coins.

In the third test case, Monocarp can buy the $2$-nd and the $3$-rd figure on the $3$-rd day, and all other figures on the $7$-th day. Then, he will spend $1+2+4+5+6 = 18$ coins.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int T,n,vis[400040];
string s;
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0),cout.tie(0);
	cin>>T;
	while(T--){
		cin>>n>>s;
		for(int i=0;i<n;i++) vis[i]=0;
		long long ans=0;
		bool ok=0;
		for(int i=n-1,j;i>0;--i)
			if(s[i]=='1' && !vis[i]){
				vis[i]=1;
				if(ok){
					while(j<i &&(s[j]=='0' || vis[j])) ++j;
					vis[j]=1;
					ans+=j+1;
//					printf("when i=%d,j=%d!\n",i+1,j+1);
					continue;
				}
				j=min(j,i-1);
				while(j>=0 &&(s[j]=='1' || vis[j])) --j;
				if(j!=-1) vis[j]=1;
				else{
					ok=1;j=0;
					while(j<i &&(s[j]=='0' || vis[j])) ++j;
					vis[j]=1;
				}
//				printf("when i=%d,j=%d!\n",i+1,j+1);
				ans+=j+1;
			}
//		printf("before:%lld\n",ans);
		for(int i=0;i<n;++i)
			if(!vis[i]) ans+=i+1;
//		printf("ans:");
		cout<<ans<<'\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:
```plaintext
4
1
1
6
101101
7
1110001
10
1101010101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Action Figures
time_limit_ms: 2500
memory_limit_mb: 512
Description: There is a shop that sells action figures near Monocarp's house. A new set of action figures will be released shortly; this set contains $n$ figures, the $i$-th figure costs $i$ coins and is available for purchase from day $i$ to day $n$.

For each of the $n$ days, Monocarp knows whether he can visit the shop.

Every time Monocarp visits the shop, he can buy any number of action figures which are sold in the shop (of course, he cannot buy an action figure that is not yet available for purchase). If Monocarp buys at least two figures during the same day, he gets a discount equal to the cost of the most expensive figure he buys (in other words, he gets the most expensive of the figures he buys for free).

Monocarp wants to buy exactly one $1$-st figure, one $2$-nd figure, ..., one $n$-th figure from the set. He cannot buy the same figure twice. What is the minimum amount of money he has to spend?

Input: The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($1 \le n \le 4 \cdot 10^5$) — the number of figures in the set (and the number of days);    * the second line contains a string $s$ ($|s| = n$, each $s_i$ is either 0 or 1). If Monocarp can visit the shop on the $i$-th day, then $s_i$ is 1; otherwise, $s_i$ is 0.

Additional constraints on the input:

  * in each test case, $s_n$ is 1, so Monocarp is always able to buy all figures during the $n$-th day;    * the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, print one integer — the minimum amount of money Monocarp has to spend.

Examples:
input:
4
1
1
6
101101
7
1110001
5
11111
output:
1
8
18
6

Note: In the first test case, Monocarp buys the $1$-st figure on the $1$-st day and spends $1$ coin.

In the second test case, Monocarp can buy the $1$-st and the $3$-rd figure on the $3$-rd day, the $2$-nd and the $4$-th figure on the $4$-th day, and the $5$-th and the $6$-th figure on the $6$-th day. Then, he will spend $1+2+5=8$ coins.

In the third test case, Monocarp can buy the $2$-nd and the $3$-rd figure on the $3$-rd day, and all other figures on the $7$-th day. Then, he will spend $1+2+4+5+6 = 18$ coins.

Buggy Code:
#pragma optimize("O3")
#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;

#define ll long long
#define ld __float128
#define ii pair<int, int>
#define pll pair<ll, ll>
#define iii array<int, 3>
#define vi vector<int>
#define vll vector<ll>
#define vii vector<ii>
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define LSOne(S) ((S) & -(S))
#define ostree tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>

void solve() {
    int n;
    cin >> n;
    string s;
    cin >> s;
    ll ans = 0;
    int oneCnt = 0;
    deque<int> zeros, ones;

    // for each value that's 0, we need to pay for it in the future
    for (int i = 0; i < n; i++) {
        if (s[i] == '0') {
            ans += i + 1;
            zeros.push_back(i);
        } else {
            ones.push_back(i);
        }
    }

    for (int i = n - 1; i >= 0; i--) {
        if (!ones.size()) break;

        if (i == ones.back()) {
            ones.pop_back();
        }
        if (s[i] == '1') {
            if (zeros.size() && zeros.back() < i) {
                // cout << "Buying " << zeros.back() << " and " << i << "\n";
                zeros.pop_back();
            } else if (ones.size() && ones.back() < i) {
                // cout << "Buying " << ones.front() << " and " << i << "\n";
                s[ones.front()] = '0';
                ans += 1 + ones.front();
                ones.pop_front();
            } else {
                // cout << "Buying " << i << "\n";
                ans += 1 + i;
            }
        }
    }
    for (int i = 0; i < ones.size(); i++) {
        ans += ones[i] + 1;
    }
    cout << ans << "\n";
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int tc = 1;
    cin >> tc;
    while (tc--) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5
10101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Action Figures
time_limit_ms: 2500
memory_limit_mb: 512
Description: There is a shop that sells action figures near Monocarp's house. A new set of action figures will be released shortly; this set contains $n$ figures, the $i$-th figure costs $i$ coins and is available for purchase from day $i$ to day $n$.

For each of the $n$ days, Monocarp knows whether he can visit the shop.

Every time Monocarp visits the shop, he can buy any number of action figures which are sold in the shop (of course, he cannot buy an action figure that is not yet available for purchase). If Monocarp buys at least two figures during the same day, he gets a discount equal to the cost of the most expensive figure he buys (in other words, he gets the most expensive of the figures he buys for free).

Monocarp wants to buy exactly one $1$-st figure, one $2$-nd figure, ..., one $n$-th figure from the set. He cannot buy the same figure twice. What is the minimum amount of money he has to spend?

Input: The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($1 \le n \le 4 \cdot 10^5$) — the number of figures in the set (and the number of days);    * the second line contains a string $s$ ($|s| = n$, each $s_i$ is either 0 or 1). If Monocarp can visit the shop on the $i$-th day, then $s_i$ is 1; otherwise, $s_i$ is 0.

Additional constraints on the input:

  * in each test case, $s_n$ is 1, so Monocarp is always able to buy all figures during the $n$-th day;    * the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, print one integer — the minimum amount of money Monocarp has to spend.

Examples:
input:
4
1
1
6
101101
7
1110001
5
11111
output:
1
8
18
6

Note: In the first test case, Monocarp buys the $1$-st figure on the $1$-st day and spends $1$ coin.

In the second test case, Monocarp can buy the $1$-st and the $3$-rd figure on the $3$-rd day, the $2$-nd and the $4$-th figure on the $4$-th day, and the $5$-th and the $6$-th figure on the $6$-th day. Then, he will spend $1+2+5=8$ coins.

In the third test case, Monocarp can buy the $2$-nd and the $3$-rd figure on the $3$-rd day, and all other figures on the $7$-th day. Then, he will spend $1+2+4+5+6 = 18$ coins.

Buggy Code:
def fi(x,y,sumi):
    if x==[]:
        print(sumi)
    elif x==[1]:
        print(sumi+1)
    elif y ==[]:
        for i in range((len(x)+1)//2):
            sumi+=x[i]
        x=[]
        fi(x,y,sumi)
    elif x[-1]-x[-2]==1:
        sumi+=y[-1]
        x.pop()
        y.pop()
        fi(x,y,sumi)
    else:
        yn=y[:]
        for i in y:
            if x[-2]<i<x[-1]:
                sumi += i
                yn.remove(i)
        x.pop()
        fi(x,yn,sumi)

x = int(input())
for i in range(x):
    y = int(input())
    z = input()
    go=[]
    notgo=[]
    for i in range(y):
        if z[i]=="1":
            go.append(int(i)+1)
        else:
            notgo.append(int(i)+1)
    fi(go,notgo,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
2
01
3
101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Action Figures
time_limit_ms: 2500
memory_limit_mb: 512
Description: There is a shop that sells action figures near Monocarp's house. A new set of action figures will be released shortly; this set contains $n$ figures, the $i$-th figure costs $i$ coins and is available for purchase from day $i$ to day $n$.

For each of the $n$ days, Monocarp knows whether he can visit the shop.

Every time Monocarp visits the shop, he can buy any number of action figures which are sold in the shop (of course, he cannot buy an action figure that is not yet available for purchase). If Monocarp buys at least two figures during the same day, he gets a discount equal to the cost of the most expensive figure he buys (in other words, he gets the most expensive of the figures he buys for free).

Monocarp wants to buy exactly one $1$-st figure, one $2$-nd figure, ..., one $n$-th figure from the set. He cannot buy the same figure twice. What is the minimum amount of money he has to spend?

Input: The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($1 \le n \le 4 \cdot 10^5$) — the number of figures in the set (and the number of days);    * the second line contains a string $s$ ($|s| = n$, each $s_i$ is either 0 or 1). If Monocarp can visit the shop on the $i$-th day, then $s_i$ is 1; otherwise, $s_i$ is 0.

Additional constraints on the input:

  * in each test case, $s_n$ is 1, so Monocarp is always able to buy all figures during the $n$-th day;    * the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, print one integer — the minimum amount of money Monocarp has to spend.

Examples:
input:
4
1
1
6
101101
7
1110001
5
11111
output:
1
8
18
6

Note: In the first test case, Monocarp buys the $1$-st figure on the $1$-st day and spends $1$ coin.

In the second test case, Monocarp can buy the $1$-st and the $3$-rd figure on the $3$-rd day, the $2$-nd and the $4$-th figure on the $4$-th day, and the $5$-th and the $6$-th figure on the $6$-th day. Then, he will spend $1+2+5=8$ coins.

In the third test case, Monocarp can buy the $2$-nd and the $3$-rd figure on the $3$-rd day, and all other figures on the $7$-th day. Then, he will spend $1+2+4+5+6 = 18$ coins.

Buggy Code:
import io
import sys
from collections import defaultdict, deque, Counter
from itertools import permutations, combinations, accumulate
from heapq import heappush, heappop
from bisect import bisect_right, bisect_left
from math import gcd
import math

_INPUT = """\
6
4
1
1
6
101101
7
1110001
5
11111
"""

def input():
  return sys.stdin.readline()[:-1]

inf=10**18
def solve(test):
  t=int(input())
  for _ in range(t):
    n=int(input())
    s=input()
    ans=n*(n+1)//2
    o,z=deque(),[]
    for i in range(n):
      if s[i]=='1':
        o.append(i+1)
      else:
        z.append(i+1)
    while o:
      x=o.pop()
      if z and z[-1]<x:
        z.pop()
        ans-=x
      elif o:
        o.popleft()
        ans-=x
    print(ans)

def random_input():
  from random import randint,shuffle
  N=randint(1,10)
  M=randint(1,N)
  A=list(range(1,M+1))+[randint(1,M) for _ in range(N-M)]
  shuffle(A)
  return (" ".join(map(str, [N,M]))+"\n"+" ".join(map(str, A))+"\n")*3

def simple_solve():
  return []

def main(test):
  if test==0:
    solve(0)
  elif test==1:
    sys.stdin = io.StringIO(_INPUT)
    case_no=int(input())
    for _ in range(case_no):
      solve(0)
  else:
    for i in range(1000):
      sys.stdin = io.StringIO(random_input())
      x=solve(1)
      y=simple_solve()
      if x!=y:
        print(i,x,y)
        print(*[line for line in sys.stdin],sep='')
        break

#0:提出用、1:与えられたテスト用、2:ストレステスト用
main(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
6
1
1
6
101101
7
1110001
5
11111
4
1101
3
101
```

### Explanation:
1. **Test Case 1:**
   - \( n = 1 \)
   - \( s = "1" \)
   - This is the simplest case where Monocarp can only buy one figure on the first day, spending 1 coin.

2. **Test Case 2:**
   - \( n = 6 \)
   - \( s = "101101" \)
   - This case is similar to one of the examples provided, where Monocarp can buy the 1st and 3rd figures on the 3rd day, the 2nd and 4th figures on the 4th day, and the 5th and 6th figures on the 6th day. The total cost is 8 coins.

3. **Test Case 3:**
   - \( n = 7 \)
   - \( s = "1110001" \)
   - This case is also similar to one of the examples provided, where Monocarp can buy the 2nd and 3rd figures on the 3rd day, and all other figures on the 7th day. The total cost is 18 coins.

4. **Test Case 4:**
   - \( n = 5 \)
   - \( s = "11111" \)
   - Monocarp can visit the shop every day, so he can buy each figure on the respective day. The total cost is 6 coins.

5. **Test Case 5:**
   - \( n = 4 \)
   - \( s = "1101" \)
   - Monocarp can buy the 1st and 2nd figures on the 2nd day, and the 3rd and 4th figures on the 4th day. The total cost is 5 coins.

6. **Test Case 6:**
   - \( n = 3 \)
   - \( s = "101" \)
   - Monocarp can buy the 1st figure on the 1st day, the 2nd and 3rd figures on the 3rd day. The total cost is 4 coins.

These test cases cover a variety of scenarios, including edge cases and typical cases, to help identify 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: Action Figures
time_limit_ms: 2500
memory_limit_mb: 512
Description: There is a shop that sells action figures near Monocarp's house. A new set of action figures will be released shortly; this set contains $n$ figures, the $i$-th figure costs $i$ coins and is available for purchase from day $i$ to day $n$.

For each of the $n$ days, Monocarp knows whether he can visit the shop.

Every time Monocarp visits the shop, he can buy any number of action figures which are sold in the shop (of course, he cannot buy an action figure that is not yet available for purchase). If Monocarp buys at least two figures during the same day, he gets a discount equal to the cost of the most expensive figure he buys (in other words, he gets the most expensive of the figures he buys for free).

Monocarp wants to buy exactly one $1$-st figure, one $2$-nd figure, ..., one $n$-th figure from the set. He cannot buy the same figure twice. What is the minimum amount of money he has to spend?

Input: The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($1 \le n \le 4 \cdot 10^5$) — the number of figures in the set (and the number of days);    * the second line contains a string $s$ ($|s| = n$, each $s_i$ is either 0 or 1). If Monocarp can visit the shop on the $i$-th day, then $s_i$ is 1; otherwise, $s_i$ is 0.

Additional constraints on the input:

  * in each test case, $s_n$ is 1, so Monocarp is always able to buy all figures during the $n$-th day;    * the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, print one integer — the minimum amount of money Monocarp has to spend.

Examples:
input:
4
1
1
6
101101
7
1110001
5
11111
output:
1
8
18
6

Note: In the first test case, Monocarp buys the $1$-st figure on the $1$-st day and spends $1$ coin.

In the second test case, Monocarp can buy the $1$-st and the $3$-rd figure on the $3$-rd day, the $2$-nd and the $4$-th figure on the $4$-th day, and the $5$-th and the $6$-th figure on the $6$-th day. Then, he will spend $1+2+5=8$ coins.

In the third test case, Monocarp can buy the $2$-nd and the $3$-rd figure on the $3$-rd day, and all other figures on the $7$-th day. Then, he will spend $1+2+4+5+6 = 18$ coins.

Buggy Code:
/* || JAI SHREE RAM || */

import java.io.*;
import java.util.*;

// Author : Sulabh Ambule

public class TLE {
  public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
  static long MOD = (long) (1e9 + 7);
  // static long MOD = 998244353;
  static FastReader in = new FastReader();

  public static void main(String[] args) throws Exception {
    int cases = in.nextInt();
    while (cases-- > 0) {
      Accepted();
    }
    out.flush();
    out.close();
  }

  private static void Accepted() {
    int n = in.nextInt();
    String s = in.next();
    int[] a = new int[n];
    for (int i = 0; i < n; i++) {
      a[i] = s.charAt(i);
    }

    long sum = n * (n + 1) / 2;
    int c = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (s.charAt(i) == '1') {
        c++;
        sum -= (i + 1);
      } else if (c > 0) {
        c--;
      }
    }
    c = Math.ceilDiv(c, 2);
    for (int i = 1; i <= c; i++) {
      sum += i;
    }
    System.out.println(sum);
  }

  static class FastReader {
    BufferedReader br;
    StringTokenizer st;

    public FastReader() {
      br = new BufferedReader(new InputStreamReader(System.in));
    }

    String next() {
      while (st == null || !st.hasMoreElements()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (IOException e) {
          e.printStackTrace();
        }
      }
      return st.nextToken();
    }

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

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

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

    String nextLine() {
      String str = "";
      try {
        str = br.readLine();
      } catch (IOException e) {
        e.printStackTrace();
      }
      return str;
    }
  }

  static class Pair {
    long first;
    long second;

    Pair(long f, long s) {
      this.first = f;
      this.second = s;
    }
  }
}

**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
4
1001
6
101011
5
11101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Action Figures
time_limit_ms: 2500
memory_limit_mb: 512
Description: There is a shop that sells action figures near Monocarp's house. A new set of action figures will be released shortly; this set contains $n$ figures, the $i$-th figure costs $i$ coins and is available for purchase from day $i$ to day $n$.

For each of the $n$ days, Monocarp knows whether he can visit the shop.

Every time Monocarp visits the shop, he can buy any number of action figures which are sold in the shop (of course, he cannot buy an action figure that is not yet available for purchase). If Monocarp buys at least two figures during the same day, he gets a discount equal to the cost of the most expensive figure he buys (in other words, he gets the most expensive of the figures he buys for free).

Monocarp wants to buy exactly one $1$-st figure, one $2$-nd figure, ..., one $n$-th figure from the set. He cannot buy the same figure twice. What is the minimum amount of money he has to spend?

Input: The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($1 \le n \le 4 \cdot 10^5$) — the number of figures in the set (and the number of days);    * the second line contains a string $s$ ($|s| = n$, each $s_i$ is either 0 or 1). If Monocarp can visit the shop on the $i$-th day, then $s_i$ is 1; otherwise, $s_i$ is 0.

Additional constraints on the input:

  * in each test case, $s_n$ is 1, so Monocarp is always able to buy all figures during the $n$-th day;    * the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, print one integer — the minimum amount of money Monocarp has to spend.

Examples:
input:
4
1
1
6
101101
7
1110001
5
11111
output:
1
8
18
6

Note: In the first test case, Monocarp buys the $1$-st figure on the $1$-st day and spends $1$ coin.

In the second test case, Monocarp can buy the $1$-st and the $3$-rd figure on the $3$-rd day, the $2$-nd and the $4$-th figure on the $4$-th day, and the $5$-th and the $6$-th figure on the $6$-th day. Then, he will spend $1+2+5=8$ coins.

In the third test case, Monocarp can buy the $2$-nd and the $3$-rd figure on the $3$-rd day, and all other figures on the $7$-th day. Then, he will spend $1+2+4+5+6 = 18$ coins.

Buggy Code:
import java.util.*;
public class actionfigures {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t>0) {
            int n=sc.nextInt();
            sc.nextLine();
            String s=sc.nextLine();
            long cost=0;
            if(n==1) {
                System.out.println(1);
                t--;
                continue;
            }
            int idx=n-1;
            while(idx>=1 && s.charAt(idx-1)=='0') {
                cost+=idx;
                idx--;
            }
            if(idx==0) {
                System.out.println(cost);
                t--;
                continue;
            }
            Deque<Integer> dq=new ArrayDeque<>();
            for(int i=1;i<=idx;i++) {
                dq.offerLast(i);
            }
            if(idx==n-1) {
                dq.offerLast(n);
            }
            while(!dq.isEmpty()) {
                cost+=dq.pollFirst();
                if(!dq.isEmpty()) {
                    dq.pollLast();
                }
            }
            System.out.println(cost);
            t--;
        }
    }
}

**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
4
3
111
4
1111
5
11111
6
111111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Action Figures
time_limit_ms: 2500
memory_limit_mb: 512
Description: There is a shop that sells action figures near Monocarp's house. A new set of action figures will be released shortly; this set contains $n$ figures, the $i$-th figure costs $i$ coins and is available for purchase from day $i$ to day $n$.

For each of the $n$ days, Monocarp knows whether he can visit the shop.

Every time Monocarp visits the shop, he can buy any number of action figures which are sold in the shop (of course, he cannot buy an action figure that is not yet available for purchase). If Monocarp buys at least two figures during the same day, he gets a discount equal to the cost of the most expensive figure he buys (in other words, he gets the most expensive of the figures he buys for free).

Monocarp wants to buy exactly one $1$-st figure, one $2$-nd figure, ..., one $n$-th figure from the set. He cannot buy the same figure twice. What is the minimum amount of money he has to spend?

Input: The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($1 \le n \le 4 \cdot 10^5$) — the number of figures in the set (and the number of days);    * the second line contains a string $s$ ($|s| = n$, each $s_i$ is either 0 or 1). If Monocarp can visit the shop on the $i$-th day, then $s_i$ is 1; otherwise, $s_i$ is 0.

Additional constraints on the input:

  * in each test case, $s_n$ is 1, so Monocarp is always able to buy all figures during the $n$-th day;    * the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, print one integer — the minimum amount of money Monocarp has to spend.

Examples:
input:
4
1
1
6
101101
7
1110001
5
11111
output:
1
8
18
6

Note: In the first test case, Monocarp buys the $1$-st figure on the $1$-st day and spends $1$ coin.

In the second test case, Monocarp can buy the $1$-st and the $3$-rd figure on the $3$-rd day, the $2$-nd and the $4$-th figure on the $4$-th day, and the $5$-th and the $6$-th figure on the $6$-th day. Then, he will spend $1+2+5=8$ coins.

In the third test case, Monocarp can buy the $2$-nd and the $3$-rd figure on the $3$-rd day, and all other figures on the $7$-th day. Then, he will spend $1+2+4+5+6 = 18$ coins.

Buggy Code:
from collections import defaultdict
import math

def solve():
    n = int(input())
    s = input()
    ans = 0

    if n <= 2:
        print(1)
        return
    if n == 3:
        print(3)
        return

    r = n-2
    if s[r] == '1':
        r = n-1

    i = 0
    while i <= r:
        if s[i] == '1' and s[r] == '1':
            ans += i+1
            i += 1
            r -= 1
        while s[r] == '0':
            ans += r+1
            r -= 1
        while s[i] == '0':
            if s[i+1] == '0':
                ans += i+1
                i += 1
            else:
                ans += i+1
                r -= 1
                i += 2
                break

    print(ans)

def main():
    for _ in range(int(input())):
        solve()

if __name__ == '__main__':
    main()

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

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

Think step by step.

Response:
```plaintext
4
1
1
6
101101
7
1110001
5
11011
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Action Figures
time_limit_ms: 2500
memory_limit_mb: 512
Description: There is a shop that sells action figures near Monocarp's house. A new set of action figures will be released shortly; this set contains $n$ figures, the $i$-th figure costs $i$ coins and is available for purchase from day $i$ to day $n$.

For each of the $n$ days, Monocarp knows whether he can visit the shop.

Every time Monocarp visits the shop, he can buy any number of action figures which are sold in the shop (of course, he cannot buy an action figure that is not yet available for purchase). If Monocarp buys at least two figures during the same day, he gets a discount equal to the cost of the most expensive figure he buys (in other words, he gets the most expensive of the figures he buys for free).

Monocarp wants to buy exactly one $1$-st figure, one $2$-nd figure, ..., one $n$-th figure from the set. He cannot buy the same figure twice. What is the minimum amount of money he has to spend?

Input: The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($1 \le n \le 4 \cdot 10^5$) — the number of figures in the set (and the number of days);    * the second line contains a string $s$ ($|s| = n$, each $s_i$ is either 0 or 1). If Monocarp can visit the shop on the $i$-th day, then $s_i$ is 1; otherwise, $s_i$ is 0.

Additional constraints on the input:

  * in each test case, $s_n$ is 1, so Monocarp is always able to buy all figures during the $n$-th day;    * the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, print one integer — the minimum amount of money Monocarp has to spend.

Examples:
input:
4
1
1
6
101101
7
1110001
5
11111
output:
1
8
18
6

Note: In the first test case, Monocarp buys the $1$-st figure on the $1$-st day and spends $1$ coin.

In the second test case, Monocarp can buy the $1$-st and the $3$-rd figure on the $3$-rd day, the $2$-nd and the $4$-th figure on the $4$-th day, and the $5$-th and the $6$-th figure on the $6$-th day. Then, he will spend $1+2+5=8$ coins.

In the third test case, Monocarp can buy the $2$-nd and the $3$-rd figure on the $3$-rd day, and all other figures on the $7$-th day. Then, he will spend $1+2+4+5+6 = 18$ coins.

Buggy Code:
t = int(input())

for tt in range(t):
    n = int(input())

    s = list(map(int, input()))
    ans = n * (n+1) // 2
    flag = 0
    for i in range(n-1, -1, -1):
        if s[i] == 0: continue
        j = i
        while j >= 0 and s[j] == 1: j -= 1

        if j == -1:
            tmp = []
            for k in range(i+1):
                if s[k] == 1: tmp.append(k+1)
            x = len(tmp)
            ans -= sum(tmp[x-(x//2):])
            break

        ans -= i+1

    print(ans)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases where the code might fail. The key aspects to consider are:

1. **Edge Cases with Small `n`**: Small values of `n` can help in understanding how the code handles simple scenarios.
2. **Edge Cases with Large `n`**: Large values of `n` can help in understanding how the code handles more complex scenarios and performance.
3. **Edge Cases with Specific Patterns in `s`**: Patterns such as long sequences of 0s or 1s, or specific positions of 1s, can help in exposing logical errors.

Given the constraints and the need for a small-scale test input, let's consider a scenario where the buggy code might fail:

### Scenario: Long Sequence of 0s Followed by 1s
In this scenario, we have a long sequence of 0s followed by a sequence of 1s. This can help in testing how the code handles the transition from days Monocarp cannot visit the shop to days he can.

### Test Input
```plaintext
1
10
0000000011
```

### Explanation
- **`n = 10`**: The number of figures and days.
- **`s = 0000000011`**: Monocarp can only visit the shop on the 9th and 10th days.

### Expected Output
- **`36`**: Monocarp can buy the 9th and 10th figures on the 9th day, and the 1st to 8th figures on the 10th day. The total cost is `1+2+3+4+5+6+7+8 = 36`.

### Why This Input is Challenging
- The long sequence of 0s ensures that Monocarp cannot buy any figures until the 9th day.
- The code needs to correctly handle the transition from 0s to 1s and ensure that the discount is applied correctly when buying multiple figures on the 9th day.
- The final day (10th) is also a valid day, but the optimal strategy is to buy the more expensive figures on the 9th day to maximize the discount.

This test input should help in exposing any logical errors 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: Action Figures
time_limit_ms: 2500
memory_limit_mb: 512
Description: There is a shop that sells action figures near Monocarp's house. A new set of action figures will be released shortly; this set contains $n$ figures, the $i$-th figure costs $i$ coins and is available for purchase from day $i$ to day $n$.

For each of the $n$ days, Monocarp knows whether he can visit the shop.

Every time Monocarp visits the shop, he can buy any number of action figures which are sold in the shop (of course, he cannot buy an action figure that is not yet available for purchase). If Monocarp buys at least two figures during the same day, he gets a discount equal to the cost of the most expensive figure he buys (in other words, he gets the most expensive of the figures he buys for free).

Monocarp wants to buy exactly one $1$-st figure, one $2$-nd figure, ..., one $n$-th figure from the set. He cannot buy the same figure twice. What is the minimum amount of money he has to spend?

Input: The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($1 \le n \le 4 \cdot 10^5$) — the number of figures in the set (and the number of days);    * the second line contains a string $s$ ($|s| = n$, each $s_i$ is either 0 or 1). If Monocarp can visit the shop on the $i$-th day, then $s_i$ is 1; otherwise, $s_i$ is 0.

Additional constraints on the input:

  * in each test case, $s_n$ is 1, so Monocarp is always able to buy all figures during the $n$-th day;    * the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, print one integer — the minimum amount of money Monocarp has to spend.

Examples:
input:
4
1
1
6
101101
7
1110001
5
11111
output:
1
8
18
6

Note: In the first test case, Monocarp buys the $1$-st figure on the $1$-st day and spends $1$ coin.

In the second test case, Monocarp can buy the $1$-st and the $3$-rd figure on the $3$-rd day, the $2$-nd and the $4$-th figure on the $4$-th day, and the $5$-th and the $6$-th figure on the $6$-th day. Then, he will spend $1+2+5=8$ coins.

In the third test case, Monocarp can buy the $2$-nd and the $3$-rd figure on the $3$-rd day, and all other figures on the $7$-th day. Then, he will spend $1+2+4+5+6 = 18$ coins.

Buggy Code:
#include <iostream>
#include <complex>
#include <queue>
#include <set>
#include <unordered_set>
#include <list>
#include <chrono>
#include <random>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <unordered_map>
#include <stack>
#include <iomanip>
#include <fstream>
using namespace std;

#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define ll long long int
#define ull unsigned long long
#define ld double
#define vi vector<int>
#define vll vector<ll>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define ff first
#define ss second
#define pb push_back
#define all(v) v.begin(),v.end()
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define inf (ll)(1e15+5)
#define mod (ll)(1e9+7)
// #define mod (ll)998244353
#define eps (ld)((1e-6))
// lower_bound(all(v),val) --- firsr value greater than or equal to val.
// upper_bound(all(v),val) --- first value greater than val.
//transform(s.begin(), s.end(), s.begin(), ::toupper);

template<class ob3>
ob3 gcd(ob3 a,ob3 b)
{   // a is greater than b (assumption)
    if(a<b) swap(a,b);
    return b == 0 ? a : gcd(b, a%b);
}
template<class ob1>
void read(vector<ob1>& v)           //reading inputs for a vector of given size.
{
    int n=(int)v.size();
    for(int i=0;i<n;i++)
        cin >> v[i];
}
template<class ob2>
void display(vector<ob2>& v)     //Displaying a vector.
{
    int n=(int)v.size();
    for(int i=0;i<n;i++)
        cout << v[i] << " ";
    cout << endl;
}

ll solve(string& s, vi& post,int dd)
{
    set<int> st;
    int n = (int)s.size();
    bool flag = false;
    int ans = 0;
    for(int i=0;i<n;i++)
    {
        flag = false;
        st.insert(i+1);
        if(s[i]=='1')
        {
            if(i==n-1) // buy all remaining
            {
                auto it = st.end();
                it--;
                int val = *(it);

                for(auto itt:st)
                {
                    ans += (ll)itt;
                    //                        cout << "itt " << itt << endl;
                }
                if(st.size()>1)
                    ans -= val;
            }
            else   // buy only two things(most expensive and least expensive)
            {
                //                    int diff = (int)st.size() - post[i+1];
                int diff = (int)st.size() - (post[i+1] - (n-(i+1)-post[i+1]));
                if(diff>dd)
                    flag = true;
                if(flag)
                {
                    auto it1 = st.begin();
                    int val1 = *(it1);
                    auto it2 = st.end();
                    it2--;
                    int val2 = *(it2);
                    st.erase(val1);
                    st.erase(val2);
                    ans += (ll)val1;
                }
            }
        }
    }
    return(ans);
}

int main()
{
    IOS;
    int tt=1;
    cin >> tt;
    while(tt--)
    {
        int n;
        cin >> n;
        string s;
        cin >> s;

        if(n==1)
        {
            cout << 1 << endl;
            continue;
        }

        ll ans = 0;

        vi post(n);
        post[n-1] = s[n-1]-'0';
        for(int i=n-2;i>=0;i--)
            post[i] = (s[i]-'0')+post[i+1];

        ans = solve(s,post,0);

        for(int i=1;i<100;i++)
        {
            ans = min(ans, solve(s,post,i));
        }

        cout << ans << endl;

    }
}

//  STAY ORGANIZED
//     .1 Have i made a mistake while executing my idea
//                 OR
//     .2 My idea is not correct
//         --    Don't get stuck on one approach.
//         --    keep edge cases in mind.
// (IMP) If getting Unexpected results, (see last point) OR try random testcases.
// CHECK :: Can the answer go beyond the INT range.
// when you get unexpected result, check for OUT_OF_BOUND and INTEGER_OVERFLOW.

**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. Here are some considerations:

1. **Edge Case with Small `n`**:
   - Test with `n = 2` or `n = 3` to see if the code handles small values correctly.

2. **Edge Case with All Visits Allowed**:
   - Test with a string where Monocarp can visit the shop every day (`s` contains all `1`s).

3. **Edge Case with No Visits Allowed Except the Last Day**:
   - Test with a string where Monocarp can visit the shop only on the last day (`s` contains all `0`s except the last character).

4. **Edge Case with Alternating Visits**:
   - Test with a string where Monocarp can visit the shop on alternating days.

5. **Edge Case with Only a Few Visits**:
   - Test with a string where Monocarp can visit the shop only a few times, but still needs to buy all figures.

Let's create a test input that combines some of these edge cases to make it challenging:

```plaintext
5
2
11
3
011
4
1011
5
00001
6
101010
```

### Explanation:
1. **Test Case 1**:
   - `n = 2`
   - `s = "11"`
   - Monocarp can visit the shop on both days. He should buy both figures on the first day for a total cost of 1 coin.

2. **Test Case 2**:
   - `n = 3`
   - `s = "011"`
   - Monocarp can visit the shop on the second and third days. He should buy the 2nd and 3rd figures on the second day and the 1st figure on the third day for a total cost of 1 + 2 = 3 coins.

3. **Test Case 3**:
   - `n = 4`
   - `s = "1011"`
   - Monocarp can visit the shop on the first, third, and fourth days. He should buy the 1st and 3rd figures on the first day, and the 2nd and 4th figures on the third day for a total cost of 1 + 2 = 3 coins.

4. **Test Case 4**:
   - `n = 5`
   - `s = "00001"`
   - Monocarp can only visit the shop on the last day. He should buy all figures on the fifth day for a total cost of 1 + 2 + 3 + 4 = 10 coins.

5. **Test Case 5**:
   - `n = 6`
   - `s = "101010"`
   - Monocarp can visit the shop on the first, third, and fifth days. He should buy the 1st and 3rd figures on the first day, the 2nd and 5th figures on the third day, and the 4th and 6th figures on the fifth day for a total cost of 1 + 2 + 4 = 7 coins.

This test input should help expose any issues in the buggy code by covering a variety of edge cases and scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Action Figures
time_limit_ms: 2500
memory_limit_mb: 512
Description: There is a shop that sells action figures near Monocarp's house. A new set of action figures will be released shortly; this set contains $n$ figures, the $i$-th figure costs $i$ coins and is available for purchase from day $i$ to day $n$.

For each of the $n$ days, Monocarp knows whether he can visit the shop.

Every time Monocarp visits the shop, he can buy any number of action figures which are sold in the shop (of course, he cannot buy an action figure that is not yet available for purchase). If Monocarp buys at least two figures during the same day, he gets a discount equal to the cost of the most expensive figure he buys (in other words, he gets the most expensive of the figures he buys for free).

Monocarp wants to buy exactly one $1$-st figure, one $2$-nd figure, ..., one $n$-th figure from the set. He cannot buy the same figure twice. What is the minimum amount of money he has to spend?

Input: The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

Each test case consists of two lines:

  * the first line contains one integer $n$ ($1 \le n \le 4 \cdot 10^5$) — the number of figures in the set (and the number of days);    * the second line contains a string $s$ ($|s| = n$, each $s_i$ is either 0 or 1). If Monocarp can visit the shop on the $i$-th day, then $s_i$ is 1; otherwise, $s_i$ is 0.

Additional constraints on the input:

  * in each test case, $s_n$ is 1, so Monocarp is always able to buy all figures during the $n$-th day;    * the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, print one integer — the minimum amount of money Monocarp has to spend.

Examples:
input:
4
1
1
6
101101
7
1110001
5
11111
output:
1
8
18
6

Note: In the first test case, Monocarp buys the $1$-st figure on the $1$-st day and spends $1$ coin.

In the second test case, Monocarp can buy the $1$-st and the $3$-rd figure on the $3$-rd day, the $2$-nd and the $4$-th figure on the $4$-th day, and the $5$-th and the $6$-th figure on the $6$-th day. Then, he will spend $1+2+5=8$ coins.

In the third test case, Monocarp can buy the $2$-nd and the $3$-rd figure on the $3$-rd day, and all other figures on the $7$-th day. Then, he will spend $1+2+4+5+6 = 18$ coins.

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

public class Codeforces
{
    public static void main(String[] args) throws java.lang.Exception
    {
        // your code goes here

        Scanner s = new Scanner(System.in);
        int t = s.nextInt();

        while (t--> 0) {
            int n = s.nextInt();
            String st = s.next();
            int str = 0;
            for (int i = n - 1; i >= 0; i--) {
                if (st.charAt(i) - '0' == 1)
                    str |= 1 << (n - i - 1);
            }
            int dis = 0;
            int leftZero = 1;
            int leftMin = 1;

            for (int i = n; i >= 0; i--) {

                if ((str & 1 << (n - i)) != 0) {
                    while (leftZero <= n - 1 && (str & 1 << leftZero) != 0) {
                        leftZero++;
                    }

                    if (leftZero <= n - 1) {
                        dis += i;
                        leftZero++;
                        if (n - leftZero < leftMin) {
                            while ((str & 1 << (n - leftMin)) == 0) leftMin++;
                        }

                    } else if (leftMin < i) {
                        leftMin++;
                        dis += i;
                        if (n - leftZero < leftMin) {
                            while ((str & 1 << (n - leftMin)) == 0) leftMin++;
                        }

                    }
                } else {
                    if (n - leftZero >= i) leftZero++;
                }

            }
            System.out.println(n * (n + 1) / 2 - dis);

        }

    }
}

**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 consider the following aspects:

1. **Edge Cases**: Test cases where the input string `s` has specific patterns that might cause the code to fail.
2. **Small Scale**: Ensure the test case is small enough to be manageable and easy to verify.
3. **Valid Input**: The input must be within the constraints specified in the problem description.

Let's analyze the code and identify potential issues:

### Potential Issues in the Buggy Code
1. **Bit Manipulation**: The code uses bit manipulation to track the days Monocarp can visit the shop. This might lead to issues if the bit manipulation logic is incorrect.
2. **Loop Conditions**: The loop conditions and the logic inside the loops might not handle edge cases correctly.
3. **Discount Calculation**: The logic for calculating the discount and the total cost might have flaws.

### Challenging Test Input
To expose the bug, we can create a test case where the string `s` has a pattern that might cause the bit manipulation or loop conditions to fail. For example, a string with a long sequence of zeros followed by a single one, or a string with a single one at the beginning and the end.

Here is a small-scale test input that should be challenging:

```plaintext
3
5
10001
6
100101
7
1001001
```

### Explanation
1. **Test Case 1**: `5 10001`
   - Monocarp can visit the shop on days 1 and 5.
   - He can buy figures 1 and 5 on day 5, and figure 2, 3, and 4 on day 1.
   - Expected output: 15 - 5 = 10

2. **Test Case 2**: `6 100101`
   - Monocarp can visit the shop on days 1, 4, and 6.
   - He can buy figures 1 and 4 on day 4, and figures 2, 3, 5, and 6 on day 6.
   - Expected output: 21 - 6 = 15

3. **Test Case 3**: `7 1001001`
   - Monocarp can visit the shop on days 1, 4, and 7.
   - He can buy figures 1 and 4 on day 4, and figures 2, 3, 5, 6, and 7 on day 7.
   - Expected output: 28 - 7 = 21

This test input should be challenging enough to expose any issues in the buggy code's logic, especially the bit manipulation and loop conditions.