**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef string st;
typedef char ch;
typedef long double ld;
#define vld vector<ld>
#define sll set<ll>
#define msll multiset<ll>
#define umll unordered_map<ll,ll>
#define mll map<ll,ll>
#define mld map<ll,ld>
#define pll pair<ll,ll>
#define pdd pair<ld,ld>
#define sch set<ch>
#define pld pair<ll,ld>
#define pdl pair<ld,ll>
#define vll vector<ll>
#define mcl map<ch,ll>
#define mcc map<ch,ch>
#define tc ll t; cin>>t; while(t--)
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
#define M 1000000007
int main(){
    tc{
        ll n;
        st s;
        cin>>n>>s;
        if(n==2){
            cout<<(ll)((s[0]-'0')*10)+s[1]-'0'<<endl;
            continue;
        }
        if(n==3){
            if((s[0]=='0')||(s[2]=='0')){
                cout<<0<<endl;
            }
            else if(s[1]=='0'){
                if(s[2]=='1'){
                    cout<<(ll)(s[0]-'0')<<endl;
                }
                else if(s[0]=='1'){
                    cout<<(ll)(s[2]-'0')<<endl;
                }
                else{
                    cout<<(ll)(s[0]+s[2]-'0'-'0')<<endl;
                }
            }
            else{
                if((s[0]=='1')&&(s[2]=='1')){
                    cout<<(ll)(10+s[1]-'0')<<endl;
                }
                else if((s[0]=='1')&&(s[1]=='1')){
                    cout<<(ll)(10+s[2]-'0')<<endl;
                }
                else if((s[1]=='1')&&(s[2]=='1')){
                    cout<<(ll)(s[0]-'0'+11)<<endl;
                }
                else{
                    if(s[0]>s[1]){
                        cout<<(ll)(s[0]-'0'+s[2]-'0'+10*(s[1]-'0'))<<endl;
                    }
                    else{
                        cout<<(ll)((s[0]-'0')*10+(s[1]-'0')+((s[2]!='1')*(s[2]-'0')))<<endl;
                    }
                }
            }
            continue;
        }
        bool x=false;
        pll u={LONG_LONG_MAX,-1};
        for(ll i=0;i<n-1;i++){
            if(s[i]=='0'){
                x=true;
                break;
            }
            if((ll)(s[i]-'0')<=u.first){
                if((u.first==(ll)(s[i]-'0'))&&(u.first!=1)){
                    continue;
                }
                u.second=i;
                u.first=(ll)(s[i]-'0');
            }
        }
        if(x){
            cout<<0<<endl;
            continue;
        }
        ll ans=0;
        for(ll j=0;j<n;j++){
            if(j==u.second){
                ans+=10*(u.first)+s[u.second+1]-'0';
                j++;
                continue;
            }
            if(s[j]!='1'){
                ans+=(s[j]-'0');
            }
        }
        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
3
3
111
4
1111
20
99999999999999999999
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

Buggy Code:
import sys
#resubmit cause weird queue
input = sys.stdin.readline

T = int(input())
for t in range(T):
    #if there is a 0 chain multiply to get 0
    #if there is a 1 multiply on one side (doesn't matter)
    #two digit number should start with 1 (so find min not at extreme)
    n = int(input())
    s = list(map(int, list(input().strip())))

    if 0 in s and len(s) > 3:
        print(0)
        continue
    if len(s) == 3:
        min_first = 1000
        min_second = 1000
        min_first = min((10 * s[0] + s[1]) + s[2], (10 * s[0] + s[1]) * s[2])
        min_first = min((10 * s[1] + s[2]) + s[0], (10 * s[1] + s[2]) * s[0])
        print(min(min_first, min_second))
        continue

    digit = 0
    for d in range(n-1):
        if s[d] < s[digit] or (s[d] ==s[digit] and s[d+1] > s[digit+1]):
            digit = d
    neuralized_ones = 0
    if s[digit] ==1:
        neuralized_ones+=1
    if s[digit+1] == 1:
        neuralized_ones += 1

    ones = sum(1 if x == 1  else 0 for x in s)
    value = sum(s) + 10 * s[digit] - s[digit] - ones +neuralized_ones#one one is used  for digit
    print(value)

**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
5
11111
6
012345
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

Buggy Code:
# 遇到0或遇到1就x，其余的全部加？
def solve():
    n = int(input())
    ss = input()
    if len(ss)<=2:return int(ss)
    s = list(ss)
    i,j = 0,1
    insertli = []
    while j < len(s) and len(insertli) < (n - 2):
        curi,curj = int(s[i]),int(s[j])
        if (s[j] == '1' or s[j] == '0') and j!=len(s) - 1:
            curj = int(s[j]+s[j+1])
            if s[j] == '0':s.pop(j)
        if curj == 0 or curi == 0:
            return 0
        if ( curi == 1 or curj == 1 ):
            insertli.append([j,'*'])
        else:
            insertli.append([j,'+'])
        i+=1
        j+=1
    step = 0
    for i in insertli:
        s.insert(i[0]+step,i[1])
        step+=1
    last = s[0]
    for i in s[1:]:
        if last == '*' and i == '*':return float('inf')
        last = i
    return min(eval(''.join(s)),solve2(n,list(ss)),solve3(n,list(ss)))

def solve2(n,s):
    if len(s) < 4:return float('inf')

    i,j = 1,2
    insertli = []
    while j < len(s) and len(insertli) < (n - 2):
        curi,curj = int(s[i]),int(s[j])
        if (s[j] == '1' or s[j] == '0') and j!=len(s) - 1:
            curj = int(s[j]+s[j+1])
            if s[j] == '0':s.pop(j)
        if curj == 0 or curi == 0:
            return 0
        if ( curi == 1 or curj == 1 ):
            insertli.append([j,'*'])
        else:
            insertli.append([j,'+'])
        i+=1
        j+=1
    step = 0
    for i in insertli:
        s.insert(i[0]+step,i[1])
        step+=1
    last = s[0]
    for i in s[1:]:
        if last == '*' and i == '*':return float('inf')
        last = i
    return eval(''.join(s))
def solve3(n,s):
    if len(s) < 4:return float('inf')
    i,j = 0,len(s) - 1
    insertli = []
    while j >=0 and len(insertli) < (n - 2):
        curi,curj = int(s[i]),int(s[j])

        if curj == 0 or curi == 0:
            return 0
        if curj == 1 :
            insertli.append([j,'*'])

        if ( curi == 1  ):
            insertli.append([i,'*'])
        i+=1
        j-=1
    step = 0
    for i in insertli:
        s.insert(i[0]+step,i[1])
        step+=1
    last = s[0]
    for i in s[1:]:
        if last == '*' and i == '*':return float('inf')
        last = i

    return eval(''.join(s))

for i in range(int(input())):print(solve())

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

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

Think step by step.

Response:
```plaintext
3
4
1111
5
20202
6
901090
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

Buggy Code:
import java.util.*;
import java.util.Comparator;
import java.util.Scanner;

public class parth {

    public static String sortString(String inputString)
    {
        // Converting input string to Character array
        Character tempArray[]
            = new Character[inputString.length()];

        for (int i = 0; i < inputString.length(); i++) {
            tempArray[i] = inputString.charAt(i);
        }

        // Sort, ignoring case during sorting
        Arrays.sort(tempArray, new Comparator<Character>() {

          // Method 2
            // To compare characters
            @Override
            public int compare(Character c1, Character c2)
            {
                // Ignoring case
                return Character.compare(
                    Character.toLowerCase(c1),
                    Character.toLowerCase(c2));
            }
        });

        // Using StringBuilder to convert Character array to
        // String
        StringBuilder sb
            = new StringBuilder(tempArray.length);

        for (Character c : tempArray)
            sb.append(c.charValue());

        return sb.toString();
    }

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t>0){
            int n= sc.nextInt();
            String s= sc.next();
            int ans = 0;
            boolean found = false;
            for(int i =0;i<n;i++){
                if(s.charAt(i) == '0'){
                    found = true;
                    break;
                }else if(s.charAt(i) == '1'){
                    ans*=(s.charAt(i)-48);
                }else if(ans==1){
                    ans*=(s.charAt(i)-48);
                }else{
                    ans+=s.charAt(i)-48;
                }
            }

            if(found){
                if(n>4){
                    System.out.println(0);
                }else if(n==2){
                    System.out.println((s.charAt(0)-48)*10 + s.charAt(1)-48);
                }else if(n==3){
                    if(s.charAt(1)=='0'){
                        if(s.charAt(0)=='1' || s.charAt(2) == '1'){
                            System.out.println(Math.max(s.charAt(0)-48,s.charAt(2)-48));
                        }else{
                            System.out.println(s.charAt(0)-48 + (s.charAt(2)-48));
                        }
                    }else{
                        System.out.println(0);
                    }
                }else{
                    System.out.println(0);
                }

            }else{
                if(n==2){
                    System.out.println((s.charAt(0)-48)*10 + s.charAt(1)-48);
                }else{
                    int fin =Integer.MAX_VALUE;
                    for(int i = 0;i<n-1;i++){
                        if(s.charAt(i)=='1' && s.charAt(i+1)=='1'){
                            int cp = ans;
                            cp+= (s.charAt(i)-48)*10 + s.charAt(i+1)-48;
                            fin = Math.min(cp,fin);

                        }else if(s.charAt(i)=='1' || s.charAt(i+1)=='1'){

                            int cp = ans-Math.max(s.charAt(i)-48,s.charAt(i+1)-48);
                            cp+= (s.charAt(i)-48)*10 + s.charAt(i+1)-48;
                            fin = Math.min(cp,fin);
                        }else{
                            int cp = ans - (s.charAt(i)-48) - (s.charAt(i+1)-48);
                            cp += (s.charAt(i)-48)*10 + s.charAt(i+1)-48;
                            fin = Math.min(cp,fin);
                        }
                    }
                    System.out.println(fin);
                }

            }
            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
5
3
101
4
1111
5
23311
6
987009
7
1111111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

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

public class mathematicalproblemthree {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

		int totaltestcases = Integer.parseInt(br.readLine());
		int[] testcasestrings = new int[totaltestcases];
		boolean goofy = false;
		for(int testcases = totaltestcases; testcases > 0; testcases--) {
			if(totaltestcases == 18) {
				int stringLength = Integer.parseInt(br.readLine());
				char[] string = br.readLine().toCharArray();

				int bestAns = Integer.MAX_VALUE;

				if(string.length > 2) {
					for(int i = 0; i < string.length - 1; i++) {
						int[] numbers = new int[string.length - 1];
						boolean doubleDigit = false;
						for(int j = 0; j < string.length; j++) {
							if(i == j) {
								numbers[j] = Character.getNumericValue(string[j]) * 10 + Character.getNumericValue(string[j + 1]);
								j++;
								doubleDigit = true;
							} else {
								if(doubleDigit) {
									numbers[j - 1] = Character.getNumericValue(string[j]);
								} else {
									numbers[j] = Character.getNumericValue(string[j]);
								}
							}
						}

						for(int j = 0; j < numbers.length; j++) {
							if(numbers[j] == 0) {
								bestAns = 0;
								continue;
							}
						}

						boolean allOne = true;
						for(int j = 0; j < numbers.length; j++) {
							if(numbers[j] != 1) {
								allOne = false;
							}
						}
						if(allOne) {
							bestAns = 1;
							continue;
						}

						int curAns = 0;
						for(int j = 0; j < numbers.length; j++) {
							if(numbers[j] != 1) {
								curAns += numbers[j];
							}
						}
						bestAns = Math.min(curAns, bestAns);
					}
				} else {
					bestAns = Integer.parseInt(new String(string));
				}
				pw.println(bestAns);
			} else {
				goofy = true;
				br.readLine();
				testcasestrings[testcases] = Integer.parseInt(br.readLine());
			}
		}

		if(goofy) {
			for(int i = 0; i < 105; i++) {
				pw.println(testcasestrings[i]);
			}
		}

		pw.close();
	}
}

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

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

Think step by step.

Response:
```plaintext
5
3
001
3
100
4
1111
4
9999
6
123456
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

Buggy Code:
def deb(*args):
    if 0:
        print("DEB:",*args)

for t in range(int(input())):
    n = int(input())
    s = input()
    a = [int(x) for x in s]
    deb("n=",n)
    deb("a=",a)
    if n==2:
        print(int(s))
    elif ( 0 in a and n>3) or (n == 3 and 0 in {a[0],a[2]}):
        print(0)
    else:
        t = [a[i]*10+a[i+1] for i in range(n-1)]
        minN = min(t)
        if minN % 10 == 1 and minN+1 in t:
            minN = minN+1
        minI = t.index(minN)
        deb("t=",t)
        deb("minN=",minN)
        deb("minI=",minI)
        a.pop(minI)
        a[minI]=minN
        a = [x for x in a if x!=1]
        deb("new a=",a)
        if(len(a)==0):
            print(1)
        else:
            print(sum(a))

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

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

Think step by step.

Response:
```plaintext
10
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

Buggy Code:
def sol(arr):
    if len(arr)==1:
        return arr[0]

    if arr==[]:
        return 0

    if len(arr)==2:
        return 10*arr[0]+arr[1]

    if 0 in arr:
        if len(arr)!=3:
            return 0
        else:
            if arr[0]==0 or arr[2]==0:
                return 0
            if arr[1]==0:
                if arr[2]==1 or arr[0]==1:
                    return arr[0]*arr[2]
                else:
                    return arr[0]+arr[2]

    mi = 100
    for i in range(len(arr)-1):
        if 10*arr[i]+arr[i+1]<mi:
            mi = 10*arr[i]+arr[i+1]
            stop = i

    while mi==11:
        if stop != len(arr)-2:
            stop+=1
        mi = 10*arr[stop]+arr[stop+1]
        if stop==len(arr)-2:
            break

    if stop == len(arr)-2 and arr[-1]==1 and arr[-2]!=1:
        if arr[-2]==arr[-3]:
            stop-=1
            mi+=1

    ans = 0
    for i in range(len(arr)):
        if i!=stop and i!=stop+1:
            if arr[i]!=1:
                ans+=arr[i]
    ans+=mi
    return ans

t = int(input())
for _ in range(t):
    n = input()
    n2 = input()
    n1 = int(n2)
    arr = []
    while n1>=1:
        arr.append(n1%10)
        n1 = n1//10
    if n2[0]=="0":
        arr.append(0)
    arr.reverse()

    print(sol(arr))

**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
3
101
4
1111
5
11111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

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

public class Main {

    public static int convert(char s){
        return s-'0';
    }

    public static void main(String[] args) {

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

        while(t-- > 0){

            int n = input.nextInt();
            input.nextLine();
            String s = input.nextLine();
            long answer = 0L;
            if(n==2){
                answer = convert(s.charAt(0)) * 10L;
                answer += convert(s.charAt(1));
                System.out.println(answer);
            }
            else{
                int index = 0;
                int min = 10;
                for(int i=0; i<n; i++){
                    if(convert(s.charAt(i))!=1)
                        answer += convert(s.charAt(i));
                    if(i == n-1 || i == 0){
                        if(convert(s.charAt(i)) == 0){
                            min = 0;
                            answer = 0;
                            break;
                        }

                    }
                    else{
                        if(i<n-1 && convert(s.charAt(i)) < min){
                            min = convert(s.charAt(i));
                            index = i;
                        }
                        else if(i<n-1 && convert(s.charAt(i)) == min){
                            if(convert(s.charAt(i+1)) != 1){
                                min = convert(s.charAt(i));
                                index = i;
                            }
                        }
                    }

                }

                if(min == 0){
                    if(n!=3)
                        answer = 0;
                    else{
                        if(index == 1 && answer == 0)
                            answer = 1;
                    }
                }
                else if(min == 1){
                    answer += 10;
                    if(convert(s.charAt(index+1)) == 1)
                        answer++;
                }
                else {
                    answer = answer - min + (min * 10);
                    if(convert(s.charAt(index+1)) == 1)
                        answer++;
                }

                System.out.println(answer);
            }

        }

    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are some considerations:

1. **Small Length Strings (n = 2)**: These should be handled correctly as no operations are performed.
2. **Strings with Leading and Trailing Zeros**: These can cause issues if the code incorrectly handles zeros.
3. **Strings with Multiple Zeros and Ones**: These can lead to incorrect handling of minimum values and index tracking.
4. **Strings with High Values**: These can help identify issues in the arithmetic operations and result calculation.
5. **Strings with Repeated Digits**: These can expose issues in the logic for finding the minimum value and its index.

Given these considerations, here is a small-scale test input that includes these edge cases:

```plaintext
5
2
10
2
01
3
011
4
1111
5
99909
```

Explanation of the test cases:
1. **2 10**: Should output `10` (no operations needed).
2. **2 01**: Should output `1` (no operations needed).
3. **3 011**: Should output `1` (since the minimum value is 0, the result should be 1).
4. **4 1111**: Should output `12` (since all digits are 1, the result should be 12).
5. **5 99909**: Should output `90` (the minimum value is 0, and the result should be 90).

This input should help identify the bugs in the given code by testing various 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: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

Buggy Code:
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Solution
{
    public static class CFScanner
    {
        BufferedReader br;
        StringTokenizer st;

        public CFScanner()
        {
            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());
        }
    }

    public static void main(String[] args)
    {
//        int n = in.nextInt();
//        long n = in.nextLong();
//        double n = in.nextDouble();
//        String s = in.next();
//        for (int i = 0; i < n; i++)

        CFScanner in = new CFScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

        int T = in.nextInt();

        for (int t = 1; t <= T; t++)
        {
            int n = in.nextInt();
            String s = in.next();
            out.println(getMinVal(n, s));
        }
        out.close();
    }

    private static int getMinVal(int n, String s)
    {
        if (s.contains("00"))
        {
            return 0;
        }
        if (n == 2)
        {
            return Integer.valueOf(s);
        }
        if (n >= 4 && s.indexOf('0') >= 0)
        {
            return 0;
        }
        int minVal = Integer.MAX_VALUE;
        int[] ds = new int[s.length()];
        for (int i = 0; i < n; i++)
        {
            ds[i] = s.charAt(i) - '0';
        }
        for (int i = 0; i <= n - 2; i++)
        {
            minVal = Math.min(minVal, getMinValHelper1(ds, i));
        }
        long[] minValHolder = new long[]{minVal};
        getMinValHelper2(ds, 1, false, minValHolder, 0L, new ArrayList<>(List.of((long)ds[0])));
        getMinValHelper2(ds, 2, true, minValHolder, 0L, new ArrayList<>(List.of((long)(10 * ds[0] + ds[1]))));
        return (int)minValHolder[0];
    }

    private static void getMinValHelper2(int[] ds, int currIdx, boolean didTwo, long[] minValHolder, long atLeastVal, List<Long> list)
    {
        if (list.get(list.size() - 1) >= minValHolder[0])
        {
            return;
        }
        if (!didTwo && currIdx == ds.length - 1)
        {
            return;
        }
        if (atLeastVal >= minValHolder[0])
        {
            return;
        }
        if (currIdx == ds.length)
        {
            if (!didTwo)
            {
                return;
            }
            minValHolder[0] = Math.min(minValHolder[0], list.stream().mapToLong(e -> e).sum());
            return;
        }
        int d = ds[currIdx];
        list.add((long)d);
        getMinValHelper2(ds, currIdx + 1, didTwo, minValHolder, atLeastVal + list.get(list.size() - 2), list);
        list.remove(list.size() - 1);
        long removed = list.remove(list.size() - 1);
        list.add((long)d * removed);
        getMinValHelper2(ds, currIdx + 1, didTwo, minValHolder, atLeastVal, list);
        list.remove(list.size() - 1);
        list.add(removed);
        if (!didTwo)
        {
            int dd = 10 * d + ds[currIdx + 1];
            list.add((long)dd);
            getMinValHelper2(ds, currIdx + 2, true, minValHolder, atLeastVal + list.get(list.size() - 2), list);
            list.remove(list.size() - 1);
            list.remove(list.size() - 1);
            list.add(dd * removed);
            getMinValHelper2(ds, currIdx + 2, true, minValHolder, atLeastVal, list);
            list.remove(list.size() - 1);
            list.add(removed);
        }
    }

    private static int getMinValHelper1(int[] ds, int idx)
    {
        int sum = 0;
        for (int i = 0; i < ds.length; i++)
        {
            if (i == idx)
            {
                continue;
            }
            sum += ds[i];
        }
        sum += 10 * ds[idx];
        return sum;
    }

    private static String getStr(String s, String c, int[] idxes)
    {
        char[] chsS = s.toCharArray();
        char[] chsC = c.toCharArray();
        int chsCIdx = 0;
        Arrays.sort(chsC);
        TreeSet<Integer> idxSet = new TreeSet<>();
        for (int idx : idxes)
        {
            idxSet.add(idx);
        }
        for (int idx : idxSet)
        {
            chsS[idx - 1] = chsC[chsCIdx++];
        }
        return new String(chsS);
    }

    private static final int[][] dirs = new int[][]{{0,1},{0,-1},{1,0},{-1,0}};

    private static void getMat(int rs, int cs, int[][] mat, PrintWriter out)
    {
        for (int r = 0; r < rs; r++)
        {
            for (int c = 0; c < cs; c++)
            {
                int curr = mat[r][c];
                int maxNei = 0;
                boolean greater = true;
                for (int[] dir : dirs)
                {
                    int rNext = r + dir[0];
                    int cNext = c + dir[1];
                    if (rNext < 0 || rNext >= rs || cNext < 0 || cNext >= cs)
                    {
                        continue;
                    }
                    maxNei = Math.max(maxNei, mat[rNext][cNext]);
                    if (mat[rNext][cNext] >= curr)
                    {
                        greater = false;
                    }
                }
                if (greater)
                {
                    mat[r][c] = maxNei;
                }
            }
        }
        for (int r = 0; r < rs; r++)
        {
            StringBuilder strBd = new StringBuilder();
            for (int c = 0; c < cs; c++)
            {
                strBd.append(mat[r][c]).append(" ");
            }
            strBd.deleteCharAt(strBd.length() - 1);
            out.println(strBd);
        }
    }

    private static int getMinDists(int x1, int x2, int x3)
    {
        List<Integer> list = new ArrayList<>(List.of(x1, x2, x3));
        Collections.sort(list);
        return list.get(2) - list.get(0);
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios that could cause the algorithm to fail. The code appears to have issues with handling zeros and the placement of symbols, especially when the string contains multiple zeros or specific patterns.

Here's a step-by-step approach to generating a small-scale test input:

1. **Test Case with Multiple Zeros**: This should trigger the early return condition `if (s.contains("00")) { return 0; }`, but we need to ensure it's not trivial and can still expose other bugs.
2. **Test Case with a Single Zero**: This should also trigger the early return condition `if (n >= 4 && s.indexOf('0') >= 0) { return 0; }`, but we need to ensure it's not too simple.
3. **Test Case with a Single Digit**: This should return the original number since no symbols can be inserted.
4. **Test Case with a String That Requires Careful Placement of Symbols**: This should test the logic for placing symbols and calculating the minimum result.

Let's create a small-scale test input based on these criteria:

```plaintext
4
4
1010
3
011
2
10
5
12345
```

Explanation:
1. **Test Case 1**: `4 1010`
   - This string contains multiple zeros and should return `0` if the early return condition works correctly. However, it also tests the logic for placing symbols.
2. **Test Case 2**: `3 011`
   - This string contains a single zero and should return `0` if the early return condition works correctly.
3. **Test Case 3**: `2 10`
   - This is a single-digit test case and should return `10` since no symbols can be inserted.
4. **Test Case 4**: `5 12345`
   - This string requires careful placement of symbols to find the minimum result, testing the main logic of the algorithm.

This test input should help expose any issues 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: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define REP(i,a,b) for (ll i=a;i<b;i++)
#define fill(a,n) REP(i,0,n){cin>>a[i];}

int main(){

ios_base::sync_with_stdio(false);
cin.tie(NULL);

#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif

int t;cin>>t;
while(t--){
    int n;
    cin>>n;
    string s;cin>>s;
    int ans=1000;
    for(int join=0;join<n-1;join++){
        int b[n-1];
        for(int i=0;i<join;i++){
            b[i]=s[i]-'0';
        }
        b[join]=(s[join]-'0')*10+(s[join+1]-'0');
        for(int i=join+1;i<n-1;i++){
            b[i]=(s[i+1]-'0');
        }
        int total_sub=(1<<(n-2));

        for(int i=0;i<total_sub;i++){
            int temp=b[0];
            for(int j=0;j<n-2;j++){
                if(i&(1<<j)){temp=temp+b[j+1];}
                else{temp=temp*b[j+1];}
                if(temp>2000)break;
            }
            ans=min(ans,temp);
        }
    }
    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
3
3
011
4
1121
5
23311
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mathematical Problem
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a string $s$ of length $n > 1$, consisting of digits from $0$ to $9$. You must insert exactly $n - 2$ symbols $+$ (addition) or $\times$ (multiplication) into this string to form a valid arithmetic expression.

In this problem, the symbols cannot be placed before the first or after the last character of the string $s$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $s = 987009$:

  * From this string, the following arithmetic expressions can be obtained: $9 \times 8 + 70 \times 0 + 9 = 81$, $98 \times 7 \times 0 + 0 \times 9 = 0$, $9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$. Note that the number $09$ is considered valid and is simply transformed into $9$.   * From this string, the following arithmetic expressions cannot be obtained: $+9 \times 8 \times 70 + 09$ (symbols should only be placed between digits), $98 \times 70 + 0 + 9$ (since there are $6$ digits, there must be exactly $4$ symbols).

The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string $s$.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. Then follows their description.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 20$) — the length of the string $s$.

The second line of each test case contains a string $s$ of length $n$, consisting of digits from $0$ to $9$.

Output: For each test case, output the minimum result of the arithmetic expression that can be obtained by inserting exactly $n - 2$ addition or multiplication symbols into the given string.

Examples:
input:
18
2
10
2
74
2
00
2
01
3
901
3
101
5
23311
6
987009
7
1111111
20
99999999999999999999
20
00000000000000000000
4
0212
18
057235283621345395
4
1112
20
19811678487321784121
4
1121
4
2221
3
011
output:
10
74
0
1
9
1
19
0
11
261
0
0
0
12
93
12
24
0

Note: In the first four test cases, we cannot add symbols, so the answer will be the original number.

In the fifth test case, the optimal answer looks as follows: $9 \times 01 = 9 \times 1 = 9$.

In the sixth test case, the optimal answer looks as follows: $1 \times 01 = 1 \times 1 = 1$.

In the seventh test case, the optimal answer looks as follows: $2 + 3 + 3 + 11 = 19$.

In the eighth test case, one of the optimal answers looks as follows: $98 \times 7 \times 0 + 0 \times 9 = 0$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
const int MOD = 1e9 + 7;

vector<vector<int>> dp;

int f(int i, int op, vector<int>& arr, int n) {
	if(i >= n) {
		if(op) return 1000;
		return 0;
	}

	if(dp[i][op] != -1) return dp[i][op];

	// int curr = arr[i];
	int curr = 1000;
	int mul = 1;
	int db = 1;

	for(int k = i; k < n; k++) {
		// if(mul > 1000) break;

		if(op && k + 1 < n) {
			int nxt = arr[k]*10 + arr[k + 1];
			int nmul = mul*nxt;

			if(nmul > 1000) break;

			curr = min(curr, nmul + f(k + 2, 0, arr, n));

			for(int l = k + 2; l < n; l++) {
				nmul *= arr[l];

				if(nmul > 1000) break;

				curr = min(curr, nmul + f(l + 1, 0, arr, n));
			}
			// db *= nxt;
			// curr = min(curr, mul*nxt + f(k + 2, 0, arr, n));
		}

		mul *= arr[k];
		if(mul > 1000) break;

		curr = min(curr, mul + f(k + 1, op, arr, n));
	}

	// cout << i << " " << curr << endl;

	return dp[i][op] = curr;
}

void solve(){
	int n; cin >> n;
	string s; cin >> s;

	vector<int> arr;
	for(auto it: s) arr.push_back(it - '0');

	dp.clear();
	dp.resize(n, vector<int>(2, -1));

	int res = f(0, 1, arr, arr.size());
	cout << res;
}

signed main() {
	ios::sync_with_stdio(false);cin.tie(NULL);

	int t; cin >> t;

	while (t--) {
		solve();
		cout << endl;
	}

	return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
2
10
3
100
4
1000
5
10000
6
100000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

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

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
#define tp tuple<int,int>
using namespace std;
const int N = 1e6+10, mod = 1e9+7;
int n, m, x, q, k,a,b;
//int a[N];
void solve(){
	cin >> n >> a >> b;
	string s;cin >> s;
	int x = 0, y = 0;
	for(int i=0;i<n;i++){
		if(s[i] == 'N')y++;
		if(s[i] == 'E')x++;
		if(s[i] == 'S')y--;
		if(s[i] == 'W')x--;
		if(x == a and y == b){
			cout << "YES\n";
			return;
		}
	}
	if(x and y and a/x == b/y){
		cout << "YES\n";
		return;
	}
	cout << "NO\n";
	cout << '\n';
}
signed main(){
	ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	int t = 1;
	cin >> t;
	while(t --)solve();
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios where the code might fail. The bug in the code lies in the logic that checks if Alice will eventually meet the Red Queen. Specifically, the code incorrectly assumes that if the ratio of the final position's coordinates matches the ratio of the Red Queen's coordinates, Alice will eventually meet the Red Queen.

Here are some edge cases to consider:
1. **Repeating Pattern that Never Reaches the Red Queen:** The pattern might not reach the Red Queen even though it repeats indefinitely.
2. **Pattern that Reaches the Red Queen After Multiple Cycles:** The pattern might reach the Red Queen after multiple cycles, which the code might not correctly handle.
3. **Pattern that Moves in a Cycle and Never Reaches the Red Queen:** The pattern might move in a cycle and never reach the Red Queen, but the code might incorrectly assume it will.

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

```plaintext
3
4 1 2
NEWS
6 1 2
NNEESW
4 2 2
NEES
```

**Explanation:**
1. **Test Case 1:**
   - Input: `n = 4, a = 1, b = 2, s = "NEWS"`
   - Alice's path: (0,0) → (0,1) → (1,1) → (1,0) → (0,0)
   - The pattern repeats indefinitely, and Alice never reaches (1,2).
   - Expected Output: `NO`

2. **Test Case 2:**
   - Input: `n = 6, a = 1, b = 2, s = "NNEESW"`
   - Alice's path: (0,0) → (0,1) → (0,2) → (1,2) → (1,1) → (0,1) → (0,0)
   - The pattern repeats indefinitely, and Alice reaches (1,2) after the first cycle.
   - Expected Output: `YES`

3. **Test Case 3:**
   - Input: `n = 4, a = 2, b = 2, s = "NEES"`
   - Alice's path: (0,0) → (0,1) → (1,1) → (1,0) → (2,0)
   - The pattern does not reach (2,2) even after multiple cycles.
   - Expected Output: `NO`

These test cases should help expose the bug in the provided code by challenging the assumptions made in the logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

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

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
// typedef unsigned int ull;
typedef long double lld;
#define Tushar                 \
   cin.sync_with_stdio(false); \
   cin.tie(0);                 \
   cout.tie(0)

// #define dbg(x) cerr << #x <<" = "; cerr<<x<<" "; cerr << endl;
#define itn long long

#define int long long int
// #define double long double
#define MAX LLONG_MAX
#define MIN LLONG_MIN
#define endl '\n'
#define vin(a)                          \
   for (int i = 0; i < (a).size(); i++) \
      cin >> a[i];
#define vout(a)                       \
   for (int i = 0; i < a.size(); i++) \
      cout << a[i] << ' ';            \
   cout << endl;
#define vpin(a)                         \
   for (int i = 0; i < (a).size(); i++) \
      cin >> a[i].first;                \
   for (int i = 0; i < (a).size(); i++) \
      cin >> a[i].second;
#define vpin2(a)                        \
   for (int i = 0; i < (a).size(); i++) \
      cin >> a[i].first >> a[i].second;
#define vpout(a)                                        \
   for (int i = 0; i < a.size(); i++)                   \
      cout << a[i].first << ' ' << a[i].second << endl; \
   cout << endl;
#define sin(a, n)              \
   for (int i = 0; i < n; i++) \
   {                           \
      int temp;                \
      cin >> temp;             \
      a.insert(temp);          \
   }
#define sout(a)         \
   for (auto i : a)     \
      cout << i << ' '; \
   cout << endl;
#define cout(x) cout << setprecision(20) << x << endl
#define isort(v) sort(v.begin(), v.end())
#define dsort(v) sort(v.begin(), v.end(), greater<int>());
#define PI 3.1415926535897932384626433832795
// #define ull unsigned long long int
// #define int int64_t
#define vi vector<int>
#define vii vector<pair<int, int>>
#define vs vector<string>
#define vc vector<char>
#define vb vector<bool>
#define pb push_back
#define loop(a, b, i) for (int i = a; i < b; i++)
#define vvi vector<vector<int>>
#define pii pair<int, int>
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define setbits(x) __builtin_popcountll(x)
#define zerobits(x) __builtin_ctzll(x)
#define r(x)             \
   {                     \
      cout << x << endl; \
      return;            \
   }
#define read(x) \
   int x;       \
   cin >> x
#define all(x) (x).begin(), (x).end()
/*---------------------------------------------------------------------------------------------------------------------------------------------*/
#ifdef ONLINE_JUDGE
#define debug(x)
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
// #include <ext/pb_ds/tree_policy.hpp>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/rope>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;

// #define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#else
#define debug(x)      \
   cerr << #x << " "; \
   _print(x);         \
   cerr << endl;
#endif

#define all(x) (x).begin(), (x).end()

#define mp make_pair
#define ff first
#define ss second

void _print(ll t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(lld t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(ull t) { cerr << t; }

template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p)
{
   cerr << "{";
   _print(p.ff);
   cerr << ",";
   _print(p.ss);
   cerr << "}";
}
template <class T>
void _print(vector<T> v)
{
   cerr << "[ ";
   for (T i : v)
   {
      _print(i);
      cerr << " ";
   }
   cerr << "]";
}
template <class T>
void _print(set<T> v)
{
   cerr << "[ ";
   for (T i : v)
   {
      _print(i);
      cerr << " ";
   }
   cerr << "]";
}
template <class T>
void _print(multiset<T> v)
{
   cerr << "[ ";
   for (T i : v)
   {
      _print(i);
      cerr << " ";
   }
   cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v)
{
   cerr << "[ ";
   for (auto i : v)
   {
      _print(i);
      cerr << " ";
   }
   cerr << "]";
}

int ok(vi & arr){
   int n = arr.size();
   int val = 1;itn val2 = 1;itn k = 1e18;
   for(itn i=0;i<n;i++){
      itn j =0;val = 1;
      while(j<n-1){
         if(j==i) j+=1;
         else {
            if(j+1!=i) {
               val = max(val,arr[j+1]-arr[j]);

               // if(i==2 and j==5){
               //    debug(arr[j+1]-arr[j])
               //    debug(val)
               // }

               j+=2;
               // debug(val)

            }
            else {
               if(j+2<n){
                  val = max(val,arr[j+2]-arr[j]);
                  j+=3;
               }
               else break;
            }
            // debug(val)
            // k =min(k,val);
            // if(k==1){
            //    debug(i)
            // }
            // if(k==1) {debug(i)debug(val)}
         }
         // if(i==2){
         //    debug(val)
         // }
         // k = min(k,val);
      }
      k = min(k,val);
   }

   return k;

   // itn l = arr[0];

   // for(itn i=1;i<n;i++){
   //    if(arr[i]-l>2*mid){
   //       l = arr[i-1];
   //       if(arr[i]-l>2*mid){
   //          return 0;
   //       }
   //    }
   // }

   // return 1;

   // // int k = min(val,val2);
   // if(k<=mid)return 1;
   // else return 0;
   // debug(k);
   // return k;

}

void try_karta_hun() {
    read(n); read(a); read(b); string s;
    cin >> s;
    int x = 0, y = 0, net_x = 0, net_y = 0;

    for (char c : s) {
        if (c == 'N') ++net_y;
        else if (c == 'E') ++net_x;
        else if (c == 'S') --net_y;
        else if (c == 'W') --net_x;
    }

    set<pair<int, int>> visited;
    for (int i = 0; i < n * n; ++i) {
        if (visited.count({x, y})) break;
        visited.insert({x, y});

        for (char c : s) {
            if (x == a && y == b) {
                yes;
                return;
            }
            if (c == 'N') ++y;
            else if (c == 'E') ++x;
            else if (c == 'S') --y;
            else if (c == 'W') --x;
        }
    }

    if (net_x == 0 && net_y == 0) {
        no;
        return;
    }

    while (true) {
        if (x == a && y == b) {
            yes;
            return;
        }
        x += net_x;
        y += net_y;

        if (abs(x) > 1e6 || abs(y) > 1e6) break;
    }
    no;
}

int32_t main(){

   Tushar;
   // precalc();
   ull t = 1;
   cin >> t;
   while (t--)
   {
      try_karta_hun();
   }

   return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases where the algorithm might fail. The code has a few potential issues:

1. **Cycle Detection**: The code uses a set to detect cycles, but it only checks for cycles within the first `n * n` iterations. This might not be sufficient if the cycle is longer.
2. **Infinite Loop Handling**: The code has a condition to break if the coordinates exceed a certain threshold (`1e6`). This might not be robust enough for all edge cases.
3. **Net Movement Calculation**: The net movement calculation is correct, but the loop that checks for meeting the Red Queen might not handle all cases where the net movement is zero.

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

1. **Cycle Detection**: Create a test case where Alice's path forms a cycle that is longer than `n * n` iterations.
2. **Infinite Loop Handling**: Create a test case where Alice's path is such that the coordinates exceed `1e6` but she can still meet the Red Queen.
3. **Net Movement Zero**: Create a test case where the net movement is zero, but Alice can still meet the Red Queen due to the cycle.

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

```plaintext
3
3 1 1
NNE
4 1 1
NSEN
5 2 2
NENWS
```

### Explanation:
1. **Test Case 1:**
   - **Input:** `3 1 1 NNE`
   - **Output:** `NO`
   - **Reason:** Alice moves to (0,1), (0,2), (1,2), and then repeats. She never reaches (1,1).

2. **Test Case 2:**
   - **Input:** `4 1 1 NSEN`
   - **Output:** `YES`
   - **Reason:** Alice moves to (0,1), (1,1), (1,0), (0,0), and then repeats. She meets the Red Queen at (1,1).

3. **Test Case 3:**
   - **Input:** `5 2 2 NENWS`
   - **Output:** `YES`
   - **Reason:** Alice moves to (0,1), (1,1), (1,2), (0,2), (0,1), and then repeats. She meets the Red Queen at (2,2).

This test input should help expose the bugs in the provided code, especially the cycle detection and infinite loop handling.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

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

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
#include <iostream>
#include <vector>
using namespace std;

bool willmeet(int n, int a, int b, string s) {
int x = 0, y = 0;
int dx = 0, dy = 0;
for (int i = 0; i < n; ++i) {
if (s[i] == 'N') y++;
if (s[i] == 'E') x++;
if (s[i] == 'S') y--;
if (s[i] == 'W') x--;
if (x == a && y == b) return true;
}
dx = x;
dy = y;
if (dx == 0 && dy == 0) return false;
if (dx != 0 && (a - 0) % dx == 0 && (b - 0) % dy == 0 && (a - 0) / dx == (b - 0) / dy) return true;
if (dy != 0 && (b - 0) % dy == 0 && (a - 0) % dx == 0 && (b - 0) / dy == (a - 0) / dx) return true;
return false;
}

int main() {
int t;
cin >> t;
while (t--) {
int n, a, b;
string s;
cin >> n >> a >> b;
cin >> s;
if (willmeet(n, a, b, s)) {
cout << "YES" << endl;
} else {
cout << "NO" << endl;
}
}
return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can highlight the issues in the algorithm. The primary issues in the buggy code are:

1. The code does not correctly handle the repeating sequence of moves.
2. The code incorrectly checks the conditions for meeting the Red Queen.

Let's break down the problem and identify a test case that can expose these issues:

1. **Repeating Sequence**: The code should consider that Alice's moves repeat indefinitely. The current implementation only checks the first cycle and then applies a incorrect condition for subsequent cycles.

2. **Incorrect Conditions**: The conditions for checking if Alice can meet the Red Queen are flawed. The code incorrectly checks the modulo conditions and the division conditions.

### Challenging Test Case

We need a test case where Alice's moves form a repeating pattern that does not immediately meet the Red Queen but could meet her after several repetitions.

### Test Case
```plaintext
3
4 3 3
NEWS
4 6 6
NEWS
4 9 9
NEWS
```

### Explanation
1. **Test Case 1**:
   - Alice starts at (0, 0) and follows the sequence "NEWS".
   - After one cycle, Alice is back at (0, 0).
   - The Red Queen is at (3, 3).
   - Since the sequence "NEWS" brings Alice back to (0, 0), Alice will never meet the Red Queen at (3, 3).
   - Expected Output: "NO"

2. **Test Case 2**:
   - Alice starts at (0, 0) and follows the sequence "NEWS".
   - After one cycle, Alice is back at (0, 0).
   - The Red Queen is at (6, 6).
   - Since the sequence "NEWS" brings Alice back to (0, 0), Alice will never meet the Red Queen at (6, 6).
   - Expected Output: "NO"

3. **Test Case 3**:
   - Alice starts at (0, 0) and follows the sequence "NEWS".
   - After one cycle, Alice is back at (0, 0).
   - The Red Queen is at (9, 9).
   - Since the sequence "NEWS" brings Alice back to (0, 0), Alice will never meet the Red Queen at (9, 9).
   - Expected Output: "NO"

### Conclusion
This test case will expose the bug in the buggy code because the code incorrectly handles the repeating sequence and the conditions for meeting the Red Queen. The expected output for all three test cases is "NO", but the buggy code might incorrectly output "YES" due to its flawed logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

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

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
import sys
input = lambda: sys.stdin.readline()[:-1]

dir = {"N": [0, 1], "E": [1, 0], "S": [0, -1], "W": [-1, 0]}

def solve():
    n, a, b = map(int, input().split())
    path = input()
    mod = [0, 0]
    vis = set()
    for c in path:
        mod[0] += dir[c][0]
        mod[1] += dir[c][1]
        vis.add(tuple(mod))
    for x, y in vis:
        A = a - x
        B = b - y
        if mod[0] == 0 and mod[1] == 0:
            if x == a and y == b:
                return "YES"
        elif mod[0] == 0:
            if A == 0 and mod[1] * (B // mod[1]) == B:
                return "YES"
        elif mod[1] == 0:
            if B == 0 and mod[0] * (A // mod[0]) == A:
                return "YES"
        elif A // mod[0] == B // mod[1]:
            return "YES"
    return "NO"

for _ in range(int(input())):
    print(solve())

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

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

Think step by step.

Response:
```plaintext
5
1 1 1
N
2 2 2
NE
3 3 3
NES
4 4 4
NEWS
5 5 5
NNEES
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

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

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
for _ in range(int(input())):
  n,x1,y1=map(int,input().split())
  s=input()
  pos=[(0,0)]
  for i in s:
    x,y=pos[-1]
    if i=="N":
      y+=1
    elif i=="S":
      y-=1
    elif i=="E":
      x+=1
    else:
      x-=1
    pos.append((x,y))
  ans="NO"
  x,y=pos[-1]
  for i in pos:
    try:
      a,b=x1-i[0],y1-i[1]
      c=0
      if x!=0 and y!=0:
        if a%x==0 and b==y*(a//x):
          ans="YES"
      elif x==0:
        if a==0 and (b==0 or b%y==0):
          ans="YES"
      else:
        if b==0 and (a==0 or a%x==0):
          ans="YES"
      if ans=="YES":
        break
    except Exception as e:
      print(str(e))
  print(ans)

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

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

Think step by step.

Response:
```plaintext
5
1 1 1
N
1 1 1
E
1 1 1
S
1 1 1
W
2 2 2
NE
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alice's Adventures in ''Chess''
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice is trying to meet up with the Red Queen in the countryside! Right now, Alice is at position $(0, 0)$, and the Red Queen is at position $(a, b)$. Alice can only move in the four cardinal directions (north, east, south, west).

More formally, if Alice is at the point $(x, y)$, she will do one of the following:

  * go north (represented by N), moving to $(x, y+1)$;    * go east (represented by E), moving to $(x+1, y)$;    * go south (represented by S), moving to $(x, y-1)$; or    * go west (represented by W), moving to $(x-1, y)$.

Alice's movements are predetermined. She has a string $s$ representing a sequence of moves that she performs from left to right. Once she reaches the end of the sequence, she repeats the same pattern of moves forever.

Can you help Alice figure out if she will ever meet the Red Queen?

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

The first line of each test case contains three integers $n$, $a$, $b$ ($1 \le n$, $a$, $b \le 10$) — the length of the string and the initial coordinates of the Red Queen.

The second line contains a string $s$ of length $n$ consisting only of the characters N, E, S, or W.

Output: For each test case, output a single string "YES" or "NO" (without the quotes) denoting whether Alice will eventually meet the Red Queen.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
2 2 2
NE
3 2 2
NNE
6 2 1
NNEESW
6 10 10
NNEESW
3 4 2
NEE
4 5 5
NEWS
output:
YES
NO
YES
YES
YES
NO

Note: In the first test case, Alice follows the path $(0,0) \xrightarrow[\texttt{N}]{} (0,1) \xrightarrow[\texttt{E}]{} (1,1) \xrightarrow[\texttt{N}]{} (1,2) \xrightarrow[\texttt{E}]{} (2,2)$.

In the second test case, Alice can never reach the Red Queen.

Buggy Code:
def solution(n, a, b, movement):
    possible = True
    x = 0
    y = 0
    if movement == '':
        return 'NO'
    while possible:
        for i in movement:
            if x == a and y == b:
                return 'YES'
            if i == 'N':
                y += 1
            elif i == 'S':
                y -= 1
            elif i == 'E':
                x += 1
            else:
                x -= 1
        if x == 0 and y == 0:
            return 'NO'
        elif x > b and y > a:
            return 'NO'

t = int(input())

for _ in range(t):
    n, a, b = map(int, input().split())
    movement = str(input())
    print(solution(n, a, b, movement))

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

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

Think step by step.

Response:
```plaintext
1
4 1 1
NEWS
```