**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating String
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako really loves alternating strings. She calls a string $s$ of lowercase Latin letters an alternating string if characters in the even positions are the same, if characters in the odd positions are the same, and the length of the string is even.

For example, the strings 'abab' and 'gg' are alternating, while the strings 'aba' and 'ggwp' are not.

As a good friend, you decided to gift such a string, but you couldn't find one. Luckily, you can perform two types of operations on the string:

  1. Choose an index $i$ and delete the $i$-th character from the string, which will reduce the length of the string by $1$. This type of operation can be performed no more than $1$ time;    2. Choose an index $i$ and replace $s_i$ with any other letter.

Since you are in a hurry, you need to determine the minimum number of operations required to make the string an alternating one.

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

The first line of each test case contains a single number $n$ ($1 \le n\le 2\cdot 10^5$)  — the length of the string.

The second line of each test case contains a string $s$, consisting of lowercase Latin letters.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the minimum number of operations required to turn the string $s$ into an alternating one.

Examples:
input:
10
1
a
2
ca
3
aab
5
ababa
6
acdada
9
ejibmyyju
6
bbccbc
6
abacba
5
bcbca
5
dcbdb
output:
1
0
1
1
2
6
2
3
1
1

Note: For the string ababa, you can delete the first character to get baba, which is an alternating string.

For the string acdada, you can change the first two characters to get dadada, which is an alternating string.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define fastio()                      \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL)
#define MOD 1000000007ll
#define tt    \
    int t;    \
    cin >> t; \
    while (t--)
#define int long long
#define pb push_back
#define all(a) a.begin(), a.end()
#define ff first
#define ss second
#define rep(i, s, e) for (int i = s; i <= e; i++)
#define repr(i, s, e) for (int i = s; i >= e; i--)
#define mp make_pair
#define vi vector<int>
#define vvi vector<vector<int>>
#define pii pair<int, int>
#define vpii vector<pii>
#define vl vector<ll>
#define vll vector<vector<ll>>
#define pll pair<ll, ll>
#define lb lower_bound
#define ub upper_bound
#define sz(a) int(a.size())
#define inf 1000000000000000000
typedef long long ll;
typedef unsigned long long ull;
typedef long double lld;
vl factorial(ll n, ll m)
{
    vl fact(n + 1, 1);
    for (int i = 2; i <= n; ++i)
    {
        fact[i] = (fact[i - 1] * i) % m;
    }
    return fact;
}
ll expo(ll a, ll b, ll mod)
{
    ll res = 1;
    while (b > 0)
    {
        if (b & 1)
            res = (res * a) % mod;
        a = (a * a) % mod;
        b = b >> 1;
    }
    return res;
}
ll mod_mul(ll a, ll b, ll m)
{
    a = a % m;
    b = b % m;
    return (((a * b) % m) + m) % m;
}
ll mminvprime(ll a, ll b) { return expo(a, b - 2, b); }
bool revsort(ll a, ll b) { return a > b; }
ll mod_div(ll a, ll b, ll m)
{
    a = a % m;
    b = b % m;
    return (mod_mul(a, mminvprime(b, m), m) + m) % m;
}
int pow_fun(int b, int p)
{
    if (p == 0)
        return 1;
    int res = 1;
    while (p)
    {
        if (p & 1)
            res *= b;
        b = b * b;
        p = p / 2;
    }
    return res;
}
vector<ll> sieve(int n)
{
    int *arr = new int[n + 1]();
    vector<ll> vect;
    for (int i = 2; i <= n; i++)
        if (arr[i] == 0)
        {
            vect.push_back(i);
            for (int j = 2 * i; j <= n; j += i)
                arr[j] = 1;
        }
    return vect;
}

int solve(int i, int potion, int turn, char a, char b, string ss, vector<vector<int>> &dp)
{
    if (i >= ss.size())
    {
        return 0;
    }
    if (dp[i][turn] != -1)
    {
        dp[i][turn];
    }

    if (i == ss.size() - 1)
    {
        if (potion)
        {
            return 1;
        }
        else
        {
            if (turn == 1)
            {
                if (ss[i] == b)
                {
                    return 0;
                }
                else
                {
                    return 1;
                }
            }
            else
            {
                if (ss[i] == a)
                {
                    return 0;
                }
                else
                {
                    return 1;
                }
            }
        }
    }

    int ans = INT_MAX;
    if (potion)
    {
        ans = min(ans, 1 + solve(i + 1, potion - 1, turn, a, b, ss, dp));
    }
    if (turn == 1)
    {
        if (ss[i] == b)
        {
            ans = min(ans, solve(i + 1, potion, !turn, a, b, ss, dp));
        }
        else
        {
            ans = min(ans, 1 + solve(i + 1, potion, !turn, a, b, ss, dp));
        }
    }
    else
    {
        if (ss[i] == a)
        {
            ans = min(ans, solve(i + 1, potion, !turn, a, b, ss, dp));
        }
        else
        {
            ans = min(ans, 1 + solve(i + 1, potion, !turn, a, b, ss, dp));
        }
    }
    return dp[i][potion] = ans;
}

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t = 0;
    cin >> t;
    while (t--)
    {
        int n = 0, m = 0, a = 0, b = 0, c = 0, d = 0, sum = 0, diff = 0, maxN = 0, minN = 0, count = 0, temp = 0;
        bool flag = false;
        cin >> n;
        string ss;
        cin >> ss;
        if (n == 1)
        {
            cout << 1 << endl;
            continue;
        }
        set<char> st;
        for (auto it : ss)
        {
            st.insert(it);
        }
        vector<char> ch;
        for (auto it : st)
        {
            ch.push_back(it);
        }

        int res = INT_MAX;
        vector<vector<int>> dp1(n, vector<int>(2, 0)), dp2(n, vector<int>(2, 0));
        for (int i = 0; i < ch.size(); i++)
        {
            char c1 = ch[i];
            char c2 = ch[i];

            if (n % 2 == 0)
            {
                int res1 = solve(0, 0, 0, c1, c2, ss, dp1);
                int res2 = solve(0, 0, 0, c2, c1, ss, dp2);
                res = min({res, res1, res2});
            }
            else
            {
                int res1 = solve(0, 1, 0, c1, c2, ss, dp1);
                int res2 = solve(0, 1, 0, c2, c1, ss, dp2);
                res = min({res, res1, res2});
            }
        }
        for (int i = 0; i < ch.size(); i++)
        {
            for (int j = 0; j < ch.size(); j++)
            {
                if (i != j)
                {
                    char c1 = ch[i];
                    char c2 = ch[j];
                    //vector<vector<int>> dp1(n, vector<int>(2, 0)), dp2(n, vector<int>(2, 0));
                    if (n % 2 == 0)
                    {
                        int res1 = solve(0, 0, 0, c1, c2, ss, dp1);
                        int res2 = solve(0, 0, 0, c2, c1, ss, dp2);
                        res = min({res, res1, res2});
                    }
                    else
                    {
                        int res1 = solve(0, 1, 0, c1, c2, ss, dp1);
                        int res2 = solve(0, 1, 0, c2, c1, ss, dp2);
                        res = min({res, res1, res2});
                    }
                }
            }
        }
        if (res == INT_MAX)
        {
            cout << 0 << endl;
            continue;
        }
        cout << res << 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
1
6
abacba
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating String
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako really loves alternating strings. She calls a string $s$ of lowercase Latin letters an alternating string if characters in the even positions are the same, if characters in the odd positions are the same, and the length of the string is even.

For example, the strings 'abab' and 'gg' are alternating, while the strings 'aba' and 'ggwp' are not.

As a good friend, you decided to gift such a string, but you couldn't find one. Luckily, you can perform two types of operations on the string:

  1. Choose an index $i$ and delete the $i$-th character from the string, which will reduce the length of the string by $1$. This type of operation can be performed no more than $1$ time;    2. Choose an index $i$ and replace $s_i$ with any other letter.

Since you are in a hurry, you need to determine the minimum number of operations required to make the string an alternating one.

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

The first line of each test case contains a single number $n$ ($1 \le n\le 2\cdot 10^5$)  — the length of the string.

The second line of each test case contains a string $s$, consisting of lowercase Latin letters.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the minimum number of operations required to turn the string $s$ into an alternating one.

Examples:
input:
10
1
a
2
ca
3
aab
5
ababa
6
acdada
9
ejibmyyju
6
bbccbc
6
abacba
5
bcbca
5
dcbdb
output:
1
0
1
1
2
6
2
3
1
1

Note: For the string ababa, you can delete the first character to get baba, which is an alternating string.

For the string acdada, you can change the first two characters to get dadada, which is an alternating string.

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

typedef unsigned int u32;
typedef long long i64;
typedef unsigned long long u64;
const double EPS = 1E-9;

const int MOD = 1E9 + 7;
const int MAXN = 200005;

void solve() {
    int n; std::cin >> n;
    std::string s; std::cin >> s; s = " " + s;
    if (n == 1) {
        std::cout << 1 << "\n";
        return;
    }
    if (n == 2) {
        std::cout << 0 << "\n";
        return;
    }
    std::vector<std::vector<std::vector<int> > > pre(n + 5, std::vector<std::vector<int> > (26, std::vector<int> (2, 0)));
    std::vector<std::vector<std::vector<int> > > suf(n + 5, std::vector<std::vector<int> > (26, std::vector<int> (2, 0)));
    for (int i = 1; i <= n; i++) {
        pre[i][s[i] - 'a'][i % 2]++;
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < 26; j++)
            for (int k = 0; k < 2; k++)
                pre[i][j][k] += pre[i - 1][j][k];
    }

    for (int i = 1; i <= n; i++) {
        suf[i][s[i] - 'a'][i % 2]++;
    }
    for (int i = n; i >= 1; --i) {
        for (int j = 0; j < 26; j++)
            for (int k = 0; k < 2; k++)
                suf[i][j][k] += suf[i + 1][j][k];
    }
    if (n & 1) {
        int mx = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < 26; j++) {
                for (int k = 0; k < 26; k++) {
                    mx = std::max(mx, pre[i - 1][j][0] + suf[i + 1][j][1] + pre[i - 1][k][1] + suf[i + 1][k][0]);
                }
            }
            // std::cout << i << " " << mx << "\n";
        }
        // for (int j = 0; j < 26; j++) for (int k = 0; k < 2; k++) std::cout << suf[7][j][k] << "\n";
        int ans = n - 1 - mx + 1;
        std::cout << ans << "\n";
    } else {
        // std::cout << "?\n";
        int mx = 0;
        for (int j = 0; j < 26; j++) {
            for (int k = 0; k < 26; k++) {
                mx = std::max(mx, pre[n][j][0] + pre[n][k][1]);
            }
        }
        int ans = n - mx;
        std::cout << ans << "\n";
    }
}

signed main() {
#ifndef ONLINE_JUDGE
freopen("IO/in", "r", stdin); freopen("IO/out", "w", stdout);
#endif
    int _ = 1; std::cin >> _;
    while (_--) { solve(); }
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
10
1
a
2
ca
3
aab
5
ababa
6
acdada
9
ejibmyyju
6
bbccbc
6
abacba
5
bcbca
5
dcbdb
```

10
1
a
2
ca
3
aab
5
ababa
6
acdada
9
ejibmyyju
6
bbccbc
6
abacba
5
bcbca
5
dcbdb


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating String
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako really loves alternating strings. She calls a string $s$ of lowercase Latin letters an alternating string if characters in the even positions are the same, if characters in the odd positions are the same, and the length of the string is even.

For example, the strings 'abab' and 'gg' are alternating, while the strings 'aba' and 'ggwp' are not.

As a good friend, you decided to gift such a string, but you couldn't find one. Luckily, you can perform two types of operations on the string:

  1. Choose an index $i$ and delete the $i$-th character from the string, which will reduce the length of the string by $1$. This type of operation can be performed no more than $1$ time;    2. Choose an index $i$ and replace $s_i$ with any other letter.

Since you are in a hurry, you need to determine the minimum number of operations required to make the string an alternating one.

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

The first line of each test case contains a single number $n$ ($1 \le n\le 2\cdot 10^5$)  — the length of the string.

The second line of each test case contains a string $s$, consisting of lowercase Latin letters.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the minimum number of operations required to turn the string $s$ into an alternating one.

Examples:
input:
10
1
a
2
ca
3
aab
5
ababa
6
acdada
9
ejibmyyju
6
bbccbc
6
abacba
5
bcbca
5
dcbdb
output:
1
0
1
1
2
6
2
3
1
1

Note: For the string ababa, you can delete the first character to get baba, which is an alternating string.

For the string acdada, you can change the first two characters to get dadada, which is an alternating string.

Buggy Code:
def test(s, a,b):
	ab = [a,b]
	n = len(s)
	if len(s)%2==0:
		return sum(s[i]!=ab[i%2] for i in range(len(s)))

	mini = len(s)
	x = sum(s[i]!=ab[1-i%2] for i in range(len(s)))
	y = 0
	for i in range(len(s)):
		x -= s[i]!=ab[1-i%2]
		mini = min(mini, y+1+x)
		y += s[i]!=ab[i%2]

	return mini

for _ in range(int(input())):
	input()
	s = input()
	mini = len(s)
	lowercase = str(set(s))
	for a in lowercase:
		for b in lowercase:
			mini = min(mini, test(s,a,b))
	print(mini)

**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
12
abacabadabac
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating String
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako really loves alternating strings. She calls a string $s$ of lowercase Latin letters an alternating string if characters in the even positions are the same, if characters in the odd positions are the same, and the length of the string is even.

For example, the strings 'abab' and 'gg' are alternating, while the strings 'aba' and 'ggwp' are not.

As a good friend, you decided to gift such a string, but you couldn't find one. Luckily, you can perform two types of operations on the string:

  1. Choose an index $i$ and delete the $i$-th character from the string, which will reduce the length of the string by $1$. This type of operation can be performed no more than $1$ time;    2. Choose an index $i$ and replace $s_i$ with any other letter.

Since you are in a hurry, you need to determine the minimum number of operations required to make the string an alternating one.

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

The first line of each test case contains a single number $n$ ($1 \le n\le 2\cdot 10^5$)  — the length of the string.

The second line of each test case contains a string $s$, consisting of lowercase Latin letters.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the minimum number of operations required to turn the string $s$ into an alternating one.

Examples:
input:
10
1
a
2
ca
3
aab
5
ababa
6
acdada
9
ejibmyyju
6
bbccbc
6
abacba
5
bcbca
5
dcbdb
output:
1
0
1
1
2
6
2
3
1
1

Note: For the string ababa, you can delete the first character to get baba, which is an alternating string.

For the string acdada, you can change the first two characters to get dadada, which is an alternating string.

Buggy Code:
import math
import sys
import itertools
from collections import Counter
from collections import deque
from collections import defaultdict
from math import gcd
from math import sqrt
from sys import stdin
import bisect
#sys.setrecursionlimit(10**7)
def input() : return stdin.readline().rstrip()
def mips():
    return map(int,input().split())
def sips():
    return map(str,input().split())
def ii():
    return int(input())
def list_perm(L) :
    return [list(l) for l in itertools.permutations(L)]
def list_perm_k(L,k):
    return [list(l) for l in itertools.permutations(L,k)]
def dist(A,B):
    x1,y1 = A
    x2,y2 = B
    return sqrt((x1-x2) ** 2 + (y1-y2) ** 2)
def ngcd(L):
    res = L[0]
    for i in range(len(L)):
        if res != 1:
            res = gcd(res,L[i])
    return res
def lcm(a,b):
    return a*b//(gcd(a,b))
def nlcm(L):
    res = L[0]
    for i in range(len(L)):
        res = lcm(res,L[i])
    return res
def ketawa(n):
    res = 0
    while (n > 0):
        res += n%10
        n //= 10
    return res

import copy
T = ii()
for _ in range(T):
    N = ii()
    s = [c for c in input()]
    prep_tables = []
    first_dict = dict()
    first_dict["odd"] = dict()
    first_dict["even"] = dict()

    for i in range(ord('a'), ord("z")+1):
        first_dict["odd"][chr(i)] = 0
        first_dict["even"][chr(i)] = 0
    for i in range(N):
        prep_tables.append(copy.deepcopy(first_dict))
    for i in range(N-1,-1,-1):
        if (i+1) % 2 == 0:
            prep_tables[i]["even"][s[i]] += 1
        else:
            prep_tables[i]["odd"][s[i]] += 1
        if i >= 1:
            prep_tables[i-1] = copy.deepcopy(prep_tables[i])
    odd = dict()
    even = dict()
    for i in range(ord('a'), ord("z")+1):
        odd[chr(i)] = 0
        even[chr(i)] = 0

    even_maxima = 0
    even_maxima_c = -1
    odd_maxima = 0
    odd_maxima_c = -1
    if N % 2 == 0:
        cnt = 0
        for i in range(N):
            if (i+1) % 2 == 0:
                even[s[i]] += 1
                if even_maxima < even[s[i]]:
                    even_maxima = even[s[i]]
                    even_maxima_c = s[i]
            else:
                odd[s[i]] += 1
                if odd_maxima < odd[s[i]]:
                    odd_maxima = odd[s[i]]
                    odd_maxima_c = s[i]
        for i in range(N):
            if (i+1) % 2 == 0 and s[i] != even_maxima_c:
                cnt += 1
            if (i+1) % 2 == 1 and s[i] != odd_maxima_c:
                cnt += 1
        print(cnt)
    if N % 2 != 0:
        odd = dict()
        even = dict()
        for i in range(ord('a'), ord("z")+1):
            odd[chr(i)] = 0
            even[chr(i)] = 0
        Evens = 0
        Odds = 0
        for i in range(N):
            if (i+1) % 2 == 0:
                even[s[i]] += 1
                Evens += 1
            else:
                odd[s[i]] += 1
                Odds += 1
        #print("before : odd = ",odd,"even = ",even)
        ans = 10**18
        for i in range(N):
            """
            #prep_tables[i]のidxが全部反転
            for key in prep_tables[i]["even"]:
                even[key] -= prep_tables[i]["even"][key]
                odd[key] += prep_tables[i]["even"][key]
            for key in prep_tables[i]["odd"]:
                even[key] += prep_tables[i]["odd"][key]
                odd[key] -= prep_tables[i]["odd"][key]
            """
            if (i+1) % 2 == 0:
                """
                even_copy = copy.deepcopy(even)
                even = copy.deepcopy(odd)
                odd = even_copy
                odd[s[i]] -= 1
                """
                new_even = copy.deepcopy(even)
                new_odd = copy.deepcopy(odd)
                new_even[s[i]] -= 1
                if i + 1 < N:
                    for key in prep_tables[i+1]["odd"]:
                        new_even[key] += prep_tables[i+1]["odd"][key]
                        new_odd[key] -= prep_tables[i+1]["odd"][key]
                    for key in prep_tables[i+1]["even"]:
                        new_odd[key] += prep_tables[i+1]["even"][key]
                        new_even[key] -= prep_tables[i+1]["even"][key]
            else:
                new_even = copy.deepcopy(even)
                new_odd = copy.deepcopy(odd)
                new_odd[s[i]] -= 1
                if i + 1 < N:
                    for key in prep_tables[i+1]["odd"]:
                        new_even[key] += prep_tables[i+1]["odd"][key]
                        new_odd[key] -= prep_tables[i+1]["odd"][key]
                    for key in prep_tables[i+1]["even"]:
                        new_odd[key] += prep_tables[i+1]["even"][key]
                        new_even[key] -= prep_tables[i+1]["even"][key]
            #print("i = ",i,"new_odd = ",new_odd,"new_even = ",new_even)
            even_maxima = -1
            odd_maxima = -1
            for key in new_even:
                if even_maxima < new_even[key]:
                    even_maxima = new_even[key]
            for key in new_odd:
                if odd_maxima < new_odd[key]:
                    odd_maxima = new_odd[key]
            #print("Evens = ",Evens,"even_maxima = ",even_maxima,"Odds = ",Odds,"odd_maxima = ",odd_maxima)
            ans = min(ans, Evens - even_maxima + Odds - odd_maxima)
        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
1
5
ababa
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating String
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako really loves alternating strings. She calls a string $s$ of lowercase Latin letters an alternating string if characters in the even positions are the same, if characters in the odd positions are the same, and the length of the string is even.

For example, the strings 'abab' and 'gg' are alternating, while the strings 'aba' and 'ggwp' are not.

As a good friend, you decided to gift such a string, but you couldn't find one. Luckily, you can perform two types of operations on the string:

  1. Choose an index $i$ and delete the $i$-th character from the string, which will reduce the length of the string by $1$. This type of operation can be performed no more than $1$ time;    2. Choose an index $i$ and replace $s_i$ with any other letter.

Since you are in a hurry, you need to determine the minimum number of operations required to make the string an alternating one.

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

The first line of each test case contains a single number $n$ ($1 \le n\le 2\cdot 10^5$)  — the length of the string.

The second line of each test case contains a string $s$, consisting of lowercase Latin letters.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the minimum number of operations required to turn the string $s$ into an alternating one.

Examples:
input:
10
1
a
2
ca
3
aab
5
ababa
6
acdada
9
ejibmyyju
6
bbccbc
6
abacba
5
bcbca
5
dcbdb
output:
1
0
1
1
2
6
2
3
1
1

Note: For the string ababa, you can delete the first character to get baba, which is an alternating string.

For the string acdada, you can change the first two characters to get dadada, which is an alternating string.

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

public class cp {
    static final FastReader in = new FastReader();
    static final FastWriter out = new FastWriter();
    static final int mod = (int) 1e9 + 7;
    static final int dirx[] = new int[] { 0, 0, 1, -1 };
    static final int diry[] = new int[] { 1, -1, 0, 0 };

    static int memo(String s, int i, char last, char prelast) {
        if (i >= s.length())
            return 0;
        int ans = (int) 1e9;

        if (s.charAt(i) == prelast) {
            ans = Math.min(ans, memo(s, i + 1, prelast, last));
        } else {
            ans = Math.min(ans, 1 + memo(s, i + 1, prelast, last));
        }
        return ans;
    }

    public static void main(String args[]) throws IOException {
        int T = in.nextInt();
        while (T-- > 0) {
            int n = in.nextInt();
            String s = in.next();
            int ans = Integer.MAX_VALUE;
            if (n % 2 == 1) {
                for (int k = 0; k < n; k++) {
                    for (char c = 'a'; c <= 'z'; c++) {
                        for (char d = 'a'; d <= 'z'; d++) {
                            ans = Math.min(ans, 1 + memo(s.substring(0, k) + s.substring(k + 1), 0, c, d));
                        }
                    }
                }
            } else {

                for (char c = 'a'; c <= 'z'; c++) {
                    for (char d = 'a'; d <= 'z'; d++) {
                        ans = Math.min(ans, memo(s, 0, c, d));
                    }
                }
            }
            out.println(ans);
        }
        out.close();
    }

    // ========================================================n================================================================================================================================================

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

    static int lcm(int a, int b) {
        return a * b / gcd(a, b);
    }

    static long pow(long n, long p, long mod) {
        if (p == 0)
            return 1;
        long next = pow(n, p / 2, mod);
        next = (next * next) % mod;
        if (p % 2 == 1l) {
            next = (next * n) % mod;
        }
        return next;
    }

    static boolean isSorted(char arr[], int i) {
        char prev = '0';
        for (char c : arr) {
            if (c < prev)
                return false;
            prev = c;
        }
        return true;
    }

    static int[] nextArr(int n) {
        int a[] = new int[(int) n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
        }
        return a;
    }

    static void reverse(int a[]) {
        for (int i = 0; i < a.length / 2; i++) {
            int tmp = a[i];
            a[i] = a[a.length - i - 1];
            a[a.length - i - 1] = tmp;
        }
    }

    static ArrayList<ArrayList<Integer>> getAdj(int n) {
        ArrayList<ArrayList<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }
        for (int i = 1; i < n; i++) {
            int u = in.nextInt() - 1;
            int v = in.nextInt() - 1;
            adj.get(u).add(v);
            adj.get(v).add(u);
        }
        return adj;
    }

    static boolean isPrime(long n) {
        if (n == 2l || n == 3l)
            return true;
        if (n == 1)
            return false;
        if (n % 2 == 0 || n % 3 == 0)
            return false;
        for (long i = 5; i * i <= n; i += 2) {
            if (n % i == 0)
                return false;
        }
        return true;
    }

    static int[] sort(int[] a) {
        ArrayList<Integer> list = new ArrayList<>();
        for (int i : a)
            list.add(i);
        Collections.sort(list);
        for (int i = 0; i < a.length; i++) {
            a[i] = list.get(i);
        }
        return a;
    }
}

class FastReader {
    BufferedReader buf = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");

    String nextToken() {
        while (!st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(buf.readLine());
            } catch (IOException e) {
            }
        }
        return st.nextToken();
    }

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

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

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

    String next() {
        try {
            return buf.readLine();
        } catch (IOException e) {
        }
        return "";
    }
}

class FastWriter {
    BufferedWriter buf = new BufferedWriter(new OutputStreamWriter(System.out));

    void print(Object object) throws IOException {
        buf.append("" + object);
    }

    void println(Object object) throws IOException {
        print(object);
        buf.append('\n');
    }

    void close() throws IOException {
        buf.close();
    }
}

class Pair {
    int ind;
    char c;

    public Pair(int ind, char c) {
        this.ind = ind;
        this.c = c;
    }

    @Override
    public String toString() {
        return "Pair [ind=" + ind + ", c=" + c + "]";
    }

    // @Override
    // public boolean equals(Object o) {
    // Pair a = (Pair) o;
    // Pair b = this;
    // return a.l == b.l && a.r == b.r && a.ind==b.ind;
    // }

    // @Override
    // public int hashCode() {
    // return ("" + this.r + this.r).hashCode();
    // }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that maximize the failure probability. The buggy code has a recursive function `memo` that does not handle all cases correctly, especially when the string length is odd.

### Step-by-Step Analysis:

1. **Odd Length Strings**:
   - The code handles odd-length strings by trying to delete one character and then applying the memo function.
   - We need to ensure that the code correctly identifies the minimum number of operations needed when the string length is odd.

2. **Even Length Strings**:
   - The code should handle even-length strings without needing to delete a character.
   - We need to ensure that the code correctly identifies the minimum number of operations needed when the string length is even.

3. **Specific Characters**:
   - The code uses a double loop to try all possible pairs of characters for even and odd positions.
   - We need to ensure that the code correctly handles cases where the string has a repeating pattern or where the characters are not alternating.

### Challenging Test Input:

Let's create a small-scale test input that includes both odd and even length strings, and strings with various patterns.

```plaintext
5
1
a
3
aba
4
abab
5
ababa
6
abacba
```

### Explanation:

1. **Test Case 1:**
   - Length: 1
   - String: "a"
   - Expected Output: 1 (Delete the only character to make an empty string, which is trivially alternating)

2. **Test Case 2:**
   - Length: 3
   - String: "aba"
   - Expected Output: 1 (Delete the middle character to get "aa", which is alternating)

3. **Test Case 3:**
   - Length: 4
   - String: "abab"
   - Expected Output: 0 (The string is already alternating)

4. **Test Case 4:**
   - Length: 5
   - String: "ababa"
   - Expected Output: 1 (Delete the first character to get "baba", which is alternating)

5. **Test Case 5:**
   - Length: 6
   - String: "abacba"
   - Expected Output: 2 (Change the third character to 'b' and the fifth character to 'a' to get "ababba", which is alternating)

This test input covers various lengths and patterns, which should help expose any bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating String
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako really loves alternating strings. She calls a string $s$ of lowercase Latin letters an alternating string if characters in the even positions are the same, if characters in the odd positions are the same, and the length of the string is even.

For example, the strings 'abab' and 'gg' are alternating, while the strings 'aba' and 'ggwp' are not.

As a good friend, you decided to gift such a string, but you couldn't find one. Luckily, you can perform two types of operations on the string:

  1. Choose an index $i$ and delete the $i$-th character from the string, which will reduce the length of the string by $1$. This type of operation can be performed no more than $1$ time;    2. Choose an index $i$ and replace $s_i$ with any other letter.

Since you are in a hurry, you need to determine the minimum number of operations required to make the string an alternating one.

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

The first line of each test case contains a single number $n$ ($1 \le n\le 2\cdot 10^5$)  — the length of the string.

The second line of each test case contains a string $s$, consisting of lowercase Latin letters.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the minimum number of operations required to turn the string $s$ into an alternating one.

Examples:
input:
10
1
a
2
ca
3
aab
5
ababa
6
acdada
9
ejibmyyju
6
bbccbc
6
abacba
5
bcbca
5
dcbdb
output:
1
0
1
1
2
6
2
3
1
1

Note: For the string ababa, you can delete the first character to get baba, which is an alternating string.

For the string acdada, you can change the first two characters to get dadada, which is an alternating string.

Buggy Code:
// Author : अभिजीत आदर्श
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class E_Alternating_String {
  static class FastReader {
    BufferedReader br;
    StringTokenizer st;

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

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

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

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

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

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

  static class FastWriter {
    private final BufferedWriter bw;

    public FastWriter() {
      this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
    }

    public void print(Object object) throws IOException {
      bw.append("" + object);
    }

    public void println(Object object) throws IOException {
      print(object);
      bw.append("\n");
    }

    public void close() throws IOException {
      bw.close();
    }
  }

  public static void main(String[] args) throws IOException {
    FastReader in = new FastReader();
    FastWriter out = new FastWriter();
    int testCases = in.nextInt();
    while (testCases-- > 0)
      blqck(in, out);
    out.close();
  }

  public static void blqck(FastReader in, FastWriter out) throws IOException {
    int n = in.nextInt();
    String s = in.nextLine();

    int[][][] freq_prefix = new int[n + 1][26][2];
    int[][][] freq_suffix = new int[n + 2][26][2];
    // [][][0] for even places

    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < 26; j++) {
        freq_prefix[i][j][0] = freq_prefix[i - 1][j][0] + ((s.charAt(i - 1) - 'a' == j && i % 2 == 0) ? 1 : 0);
        freq_prefix[i][j][1] = freq_prefix[i - 1][j][1] + ((s.charAt(i - 1) - 'a' == j && i % 2 == 1) ? 1 : 0);
      }
    }
    for (int i = n; i >= 1; i--) {
      for (int j = 0; j < 26; j++) {
        freq_suffix[i][j][0] = freq_suffix[i + 1][j][0] + ((s.charAt(i - 1) - 'a' == j && i % 2 == 0) ? 1 : 0);
        freq_suffix[i][j][1] = freq_suffix[i + 1][j][1] + ((s.charAt(i - 1) - 'a' == j && i % 2 == 1) ? 1 : 0);
      }
    }

    int ans = 2 * n;
    if (s.length() % 2 == 0) {
      // no need to remove a char
      int[] max_freq = new int[2];
      for (int i = 0; i < 26; i++) {
        max_freq[0] = Math.max(max_freq[0], freq_suffix[1][i][0]);
        max_freq[1] = Math.max(max_freq[1], freq_suffix[1][i][1]);
      }
      ans = n - max_freq[0] - max_freq[1];
    } else {
      for (int i = 1; i <= n; i++) {
        int prev = i - 1, next = i + 1;
        int[] max_freq = new int[2];
        for (int j = 0; j < 26; j++) {
          max_freq[0] = Math.max(max_freq[0], freq_prefix[prev][j][0] + freq_suffix[next][j][1]);
          max_freq[1] = Math.max(max_freq[1], freq_prefix[prev][j][1] + freq_suffix[next][j][0]);
        }
        ans = Math.min(ans, n - max_freq[0] - max_freq[1]);
      }
    }

    out.println(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
1
7
abacaba
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating String
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako really loves alternating strings. She calls a string $s$ of lowercase Latin letters an alternating string if characters in the even positions are the same, if characters in the odd positions are the same, and the length of the string is even.

For example, the strings 'abab' and 'gg' are alternating, while the strings 'aba' and 'ggwp' are not.

As a good friend, you decided to gift such a string, but you couldn't find one. Luckily, you can perform two types of operations on the string:

  1. Choose an index $i$ and delete the $i$-th character from the string, which will reduce the length of the string by $1$. This type of operation can be performed no more than $1$ time;    2. Choose an index $i$ and replace $s_i$ with any other letter.

Since you are in a hurry, you need to determine the minimum number of operations required to make the string an alternating one.

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

The first line of each test case contains a single number $n$ ($1 \le n\le 2\cdot 10^5$)  — the length of the string.

The second line of each test case contains a string $s$, consisting of lowercase Latin letters.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the minimum number of operations required to turn the string $s$ into an alternating one.

Examples:
input:
10
1
a
2
ca
3
aab
5
ababa
6
acdada
9
ejibmyyju
6
bbccbc
6
abacba
5
bcbca
5
dcbdb
output:
1
0
1
1
2
6
2
3
1
1

Note: For the string ababa, you can delete the first character to get baba, which is an alternating string.

For the string acdada, you can change the first two characters to get dadada, which is an alternating string.

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

bool cmp(std::pair<char, int> a, std::pair<char, int> b) {
    return a.second > b.second;
}

int count_other_letters(std::string s) {
    int n = s.size();
    std::map<char, int> count_even;
    std::map<char, int> count_odd;
    char c;
    for (int i = 0; i < n; i++) {
        c = s[i];
        if (i % 2 == 0) {
            count_even[c]++;
        }
        else {
            count_odd[c]++;
        }
    }

    std::vector<std::pair<char, int>> even_v;
    std::vector<std::pair<char, int>> odd_v;

    for (auto p : count_even) {
        even_v.push_back(p);
    }

    for (auto p : count_odd) {
        odd_v.push_back(p);
    }

    std::sort(even_v.begin(), even_v.end(), cmp);
    std::sort(odd_v.begin(), odd_v.end(), cmp);

    int other = n;
    if (even_v.size() > 0) {
        other -= even_v[0].second;
    }
    if (odd_v.size() > 0) {
        other -= odd_v[0].second;
    }
    return other;
}

void solve() {
    int n;
    std::cin >> n;

    std::string s;
    std::cin >> s;

    int ans = count_other_letters(s);
    if (n % 2 == 0) {
        std::cout << ans << std::endl;
    }
    else {
        ans = 10e8;
        std::string copy;
        for (int i = 0; i < n; i++) {
            copy = s;
            copy.erase(copy.begin() + i);
            int count = count_other_letters(copy);
            if (count + 1< ans) {
                ans = count + 1;
            }
        }
            std::cout << ans << std::endl;
    }

}

int main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(NULL);
    long long t;
    std::cin >> t;
    for (long long i = 0; i < t; i++) {
        solve();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases that can highlight issues in the logic. Here are some considerations:

1. **Odd-length strings**: The code handles odd-length strings by removing one character and then checking the minimum operations required. We should test with odd-length strings to ensure the logic for removing a character and recalculating is correct.
2. **Strings with repeated characters**: These can cause issues if the code does not correctly handle the most frequent characters in even and odd positions.
3. **Strings with alternating patterns**: These should require zero operations, and the code should handle them correctly.
4. **Strings that are close to being alternating but require minimal changes**: These can help test the logic for replacing characters.

Let's create a small-scale test input that includes these edge cases:

```plaintext
5
3
aab
5
ababa
6
acdada
9
ejibmyyju
7
abcabca
```

Explanation:
- **3 aab**: This is an odd-length string that should require 1 operation to become alternating.
- **5 ababa**: This is an odd-length string that can be made alternating by removing the first character.
- **6 acdada**: This string can be made alternating by changing the first two characters.
- **9 ejibmyyju**: This string requires 6 operations to become alternating.
- **7 abcabca**: This is an odd-length string that should require 3 operations to become alternating.

This test input should help expose any issues in the buggy code.

5
3
aab
5
ababa
6
acdada
9
ejibmyyju
7
abcabca


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating String
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako really loves alternating strings. She calls a string $s$ of lowercase Latin letters an alternating string if characters in the even positions are the same, if characters in the odd positions are the same, and the length of the string is even.

For example, the strings 'abab' and 'gg' are alternating, while the strings 'aba' and 'ggwp' are not.

As a good friend, you decided to gift such a string, but you couldn't find one. Luckily, you can perform two types of operations on the string:

  1. Choose an index $i$ and delete the $i$-th character from the string, which will reduce the length of the string by $1$. This type of operation can be performed no more than $1$ time;    2. Choose an index $i$ and replace $s_i$ with any other letter.

Since you are in a hurry, you need to determine the minimum number of operations required to make the string an alternating one.

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

The first line of each test case contains a single number $n$ ($1 \le n\le 2\cdot 10^5$)  — the length of the string.

The second line of each test case contains a string $s$, consisting of lowercase Latin letters.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the minimum number of operations required to turn the string $s$ into an alternating one.

Examples:
input:
10
1
a
2
ca
3
aab
5
ababa
6
acdada
9
ejibmyyju
6
bbccbc
6
abacba
5
bcbca
5
dcbdb
output:
1
0
1
1
2
6
2
3
1
1

Note: For the string ababa, you can delete the first character to get baba, which is an alternating string.

For the string acdada, you can change the first two characters to get dadada, which is an alternating string.

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

using namespace std;

#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
struct custom_hash {
static uint64_t splitmix64(uint64_t x) {
    // http://xorshift.di.unimi.it/splitmix64.c
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
return splitmix64(x + FIXED_RANDOM);
}
};
 typedef long long ll;
 #define vec vector<ll>
 #define str string
#define matrix vector<vec>
//# this one is specailly for multiset
typedef tree<
pair<ll, ll>,
null_type,
less<pair<ll, ll>>,
rb_tree_tag,
tree_order_statistics_node_update> ordered_set;
 //comment this out if you do not want to use order_of_key(elements strictly lesser in multiset
// int t = 0;
// ordered_set me;
// ...
// me.insert({x, t++});
// me.erase(me.lower_bound({x, 0}));
// cout << me.order_of_key({x, 0}) <<
;
ll sum(vec &arr){ ll num=0; for(auto it:arr){num+=it; } return num;  }
bool compare1(const vec&v1,const vec&v2){ return ( v1[0]<v2[0]); }
bool compare2(const vec&v1,const vec&v2){ return (v1[0]>v2[0]); }
vec prefix(vec &arr){ ll sum=0; vec answer; for(auto it:arr){ sum+= it; answer.push_back(sum); } return answer; }
 ll gcdExtended(ll a, ll b, ll &x, ll &y) {
if (a == 0) {
x = 0;
y = 1;
return b;
}ll x1, y1;
ll gcd = gcdExtended(b % a, a, x1, y1);
x = y1 - (b / a) * x1;y = x1;
return gcd;}
ll modInverse(ll a, ll m) {
ll x, y;
ll gcd = gcdExtended(a, m, x, y);
if (gcd != 1) {return -1;}
return (x % m + m) % m;}

ll solve1(str s){
    ll n= s.size();
    map<char,ll>mp1,mp2;
    for(ll i{0};i<s.size();i++){
        if(i%2==0){
            mp1[s[i]]++;
        }
        else{
            mp2[s[i]]++;
        }
    }

    ll maxi=INT_MIN;
    for(auto it:mp1){ maxi=max(maxi,it.second); }
    ll temp = ( n/2) - maxi;
    maxi=INT_MIN;
    for(auto it:mp2){ maxi= max(maxi,it.second); }
    temp= temp+ ( n/2) - maxi;
    return temp;
}

ll solve2(str s){

ll n= s.size();
vector< map<char,ll> > even_right(n);
vector< map<char,ll> > odd_right(n);
vector< map<char,ll> > even_left(n);
vector< map<char,ll> > odd_left(n);

map<char,ll>mp1,mp2;

for(ll i{0};i<n;i++){
    odd_left[i]=mp1;
    even_left[i]=mp2;
    if(i%2==0){
        mp1[s[i]]++;
    }
    else{
        mp2[s[i]]++;
    }
}

map<char,ll>mp3,mp4;

for(ll i{n-1};i>=0;i--){
     odd_right[i]=mp3;
     even_right[i]=mp4;
    if(i%2==0){
        mp3[s[i]]++;
    }
    else{
        mp4[s[i]]++;
    }
}

ll answer=INT_MAX;

for(ll i{0};i<n;i++){
    if(i==0){
        ll maxi=INT_MIN;
        for(auto it:odd_right[i]){ maxi=max(maxi,it.second);}
        ll temp= (n-1)/2; temp= temp-maxi;
        maxi= INT_MIN;
        for(auto it:even_right[i]){ maxi=max(maxi,it.second);  }
        temp= temp + ( (n-1)/2 ) - maxi;
        answer=min(answer,temp);
        continue;
    }
      if(i==n-1){
        ll maxi=INT_MIN;
        for(auto it:odd_left[i]){ maxi=max(maxi,it.second);}
        ll temp= (n-1)/2; temp= temp-maxi;
        maxi= INT_MIN;
        for(auto it:even_left[i]){ maxi=max(maxi,it.second);  }
        temp= temp + ( (n-1)/2 ) - maxi;
        answer=min(answer,temp);
        continue;
    }

     map< char,ll> idk1,idk2;
     for(auto it:odd_left[i]){ idk1[it.first]+= it.second; }
     for(auto it:even_right[i]){ idk1[it.first]+= it.second; }
     for(auto it:even_left[i]){ idk2[it.first]+= it.second; }
     for(auto it:odd_right[i]){ idk2[it.first]+= it.second; }

     ll maxi= INT_MIN;
     for(auto it:idk1){ maxi= max(maxi,it.second); }
     ll temp= (n-1)/2; temp= temp-maxi;
     maxi=INT_MIN;
     for(auto it:idk2){ maxi= max(maxi,it.second); }
     temp= temp+ ( (n-1)/2) - maxi;
     answer= min(answer, temp);
}

return answer+1;
}

int main(){

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

ll tc; cin >> tc;

while(tc--){
    ll n; cin >> n;
    str s; cin >> s;
    if(n==1){cout << 1 << endl; continue; }

    if(n%2==0){
        cout << solve1(s) << endl; continue;
    }

    cout << solve2(s) << 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
1
a
2
ca
3
aab
5
ababa
6
acdada
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating String
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako really loves alternating strings. She calls a string $s$ of lowercase Latin letters an alternating string if characters in the even positions are the same, if characters in the odd positions are the same, and the length of the string is even.

For example, the strings 'abab' and 'gg' are alternating, while the strings 'aba' and 'ggwp' are not.

As a good friend, you decided to gift such a string, but you couldn't find one. Luckily, you can perform two types of operations on the string:

  1. Choose an index $i$ and delete the $i$-th character from the string, which will reduce the length of the string by $1$. This type of operation can be performed no more than $1$ time;    2. Choose an index $i$ and replace $s_i$ with any other letter.

Since you are in a hurry, you need to determine the minimum number of operations required to make the string an alternating one.

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

The first line of each test case contains a single number $n$ ($1 \le n\le 2\cdot 10^5$)  — the length of the string.

The second line of each test case contains a string $s$, consisting of lowercase Latin letters.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the minimum number of operations required to turn the string $s$ into an alternating one.

Examples:
input:
10
1
a
2
ca
3
aab
5
ababa
6
acdada
9
ejibmyyju
6
bbccbc
6
abacba
5
bcbca
5
dcbdb
output:
1
0
1
1
2
6
2
3
1
1

Note: For the string ababa, you can delete the first character to get baba, which is an alternating string.

For the string acdada, you can change the first two characters to get dadada, which is an alternating string.

Buggy Code:
from collections import Counter

for _ in range(int(input())):
    ans = 0
    n = int(input())
    s = input()
    def solve(s):
        n = len(s)
        ans = 0
        a = Counter([s[i] for i in range(0,n,2)])
        b = Counter([s[i] for i in range(1,n,2)])
        k = 9e9
        for key in a:
            k = min(k,n//2-a[key])
        ans += k
        k = 9e9
        for key in b:
            k = min(k,n//2-b[key])
        ans += k
        return ans

    if len(s)==1:
        print(1)
    elif len(s)%2:
        ans = 9e9
        for i in range(n):
            ans = min(ans,solve(s[0:i]+s[i+1:]))
        print(ans+1)
    else:
        print(solve(s))

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

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

Think step by step.

Response:
```plaintext
10
1
a
2
ca
3
aab
5
ababa
6
acdada
9
ejibmyyju
6
bbccbc
6
abacba
5
bcbca
5
dcbdb
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating String
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako really loves alternating strings. She calls a string $s$ of lowercase Latin letters an alternating string if characters in the even positions are the same, if characters in the odd positions are the same, and the length of the string is even.

For example, the strings 'abab' and 'gg' are alternating, while the strings 'aba' and 'ggwp' are not.

As a good friend, you decided to gift such a string, but you couldn't find one. Luckily, you can perform two types of operations on the string:

  1. Choose an index $i$ and delete the $i$-th character from the string, which will reduce the length of the string by $1$. This type of operation can be performed no more than $1$ time;    2. Choose an index $i$ and replace $s_i$ with any other letter.

Since you are in a hurry, you need to determine the minimum number of operations required to make the string an alternating one.

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

The first line of each test case contains a single number $n$ ($1 \le n\le 2\cdot 10^5$)  — the length of the string.

The second line of each test case contains a string $s$, consisting of lowercase Latin letters.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the minimum number of operations required to turn the string $s$ into an alternating one.

Examples:
input:
10
1
a
2
ca
3
aab
5
ababa
6
acdada
9
ejibmyyju
6
bbccbc
6
abacba
5
bcbca
5
dcbdb
output:
1
0
1
1
2
6
2
3
1
1

Note: For the string ababa, you can delete the first character to get baba, which is an alternating string.

For the string acdada, you can change the first two characters to get dadada, which is an alternating string.

Buggy Code:
def get_operations(s):
    odds = [s[i] for i in range(0, len(s), 2)]
    odd_dict = {i: 0 for i in odds}
    for i in odds:
        odd_dict[i] += 1
    odds_operations = len(odds) - max(odd_dict.values())

    evens = [s[i] for i in range(1, len(s), 2)]
    even_dict = {i: 0 for i in evens}
    for i in evens:
        even_dict[i] += 1
    evens_operations = len(evens) - max(even_dict.values())
    return odds_operations+evens_operations

if __name__ == "__main__":
    t = int(input())
    for _ in range(t):
        n = int(input())
        s = input()
        if n == 1:
            print(1)
        elif n == 2:
            print(0)
        else:
            need_deleted = True if n % 2 == 1 else False
            if need_deleted is False:
                ops = get_operations(s)
                print(ops)
            else:
                min_ops = float("inf")
                for i in range(len(s)):
                    temp = s[:i] + s[i+1:]
                    ops = get_operations(temp)
                    min_ops = min(min_ops, ops)
                print(min_ops+1)

**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
7
abacaba
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating String
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako really loves alternating strings. She calls a string $s$ of lowercase Latin letters an alternating string if characters in the even positions are the same, if characters in the odd positions are the same, and the length of the string is even.

For example, the strings 'abab' and 'gg' are alternating, while the strings 'aba' and 'ggwp' are not.

As a good friend, you decided to gift such a string, but you couldn't find one. Luckily, you can perform two types of operations on the string:

  1. Choose an index $i$ and delete the $i$-th character from the string, which will reduce the length of the string by $1$. This type of operation can be performed no more than $1$ time;    2. Choose an index $i$ and replace $s_i$ with any other letter.

Since you are in a hurry, you need to determine the minimum number of operations required to make the string an alternating one.

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

The first line of each test case contains a single number $n$ ($1 \le n\le 2\cdot 10^5$)  — the length of the string.

The second line of each test case contains a string $s$, consisting of lowercase Latin letters.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the minimum number of operations required to turn the string $s$ into an alternating one.

Examples:
input:
10
1
a
2
ca
3
aab
5
ababa
6
acdada
9
ejibmyyju
6
bbccbc
6
abacba
5
bcbca
5
dcbdb
output:
1
0
1
1
2
6
2
3
1
1

Note: For the string ababa, you can delete the first character to get baba, which is an alternating string.

For the string acdada, you can change the first two characters to get dadada, which is an alternating string.

Buggy Code:
import java.util.*;
public class Selection_sort {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int tc = sc.nextInt();
        while(tc-->0) {
            int n = sc.nextInt();
            String str = sc.next();
            if(n==1){
                System.out.println(1);
                continue;
            }
            if (n % 2 != 0) {
                ArrayList<int[][]> left = new ArrayList<>();
                for (int i = 0; i < n; i++) {
                    left.add(new int[26][26]);
                    if (i % 2 == 0) {
                        left.get(i)[0][str.charAt(i) - 'a']++;
                    } else {
                        left.get(i)[1][str.charAt(i) - 'a']++;
                    }
                    if (i != 0) {
                        for (int j = 0; j < 26; j++) {
                            left.get(i)[0][j] += left.get(i - 1)[0][j];
                            left.get(i)[1][j] += left.get(i - 1)[1][j];
                        }
                    }
                }
                ArrayList<int[][]> right = new ArrayList<>();
                for (int i = n - 1; i >= 0; i--) {
                    right.add(new int[26][26]);
                    if (i % 2 == 0) {
                        right.get(n - 1 - i)[0][str.charAt(i) - 'a']++;
                    } else {
                        right.get(n - 1 - i)[1][str.charAt(i) - 'a']++;
                    }
                    if (n - 1 - i != 0) {
                        for (int j = 0; j < 26; j++) {
                            right.get(n - 1 - i)[0][j] += right.get(n - 1 - 1 - i)[0][j];
                            right.get(n - 1 - i)[1][j] += right.get(n - 1 - 1 - i)[1][j];
                        }
                    }
                }
                Collections.reverse(right);
                int ans = Integer.MAX_VALUE;
                for (int i = 0; i < n; i++) {
                    int checkEven[] = new int[26];
                    int checkOdd[] = new int[26];
                    if(i%2==0){
                        left.get(i)[0][str.charAt(i)-'a']--;
                        right.get(i)[0][str.charAt(i)-'a']--;
                    }else{
                        left.get(i)[1][str.charAt(i)-'a']--;
                        right.get(i)[1][str.charAt(i)-'a']--;
                    }
                    for (int j = 0; j < 26; j++) {
                        checkEven[j] = left.get(i)[0][j];
                        checkOdd[j] = left.get(i)[1][j];
                    }
                    for (int j = 0; j < 26; j++) {
                            checkEven[j] += right.get(i)[1][j];
                            checkOdd[j] += right.get(i)[0][j];
                    }

                    int maxEven = 0;
                    int maxOdd = 0;
                    for (int j = 0; j < 26; j++) {
                        maxEven = Math.max(maxEven, checkEven[j]);
                        maxOdd = Math.max(maxOdd, checkOdd[j]);
                    }
                    int x = n;
                    x--;
                    ans = Math.min(x / 2 - maxEven + x / 2 - maxOdd + 1, ans);
                    if(i%2==0){
                        left.get(i)[0][str.charAt(i)-'a']++;
                        right.get(i)[0][str.charAt(i)-'a']++;
                    }else{
                        left.get(i)[1][str.charAt(i)-'a']++;
                        right.get(i)[1][str.charAt(i)-'a']++;
                    }
                }
                System.out.println(ans);
            }else{
                int even[]=new int[26];
                int odd[]=new int[26];
                int maxEven=0;
                int maxOdd=0;
                for(int i=0;i<n;i++){
                    if(i%2==0){
                        even[str.charAt(i)-'a']++;
                    }else{
                        odd[str.charAt(i)-'a']++;
                    }
                    maxEven = Math.max(maxEven,even[str.charAt(i)-'a']);
                    maxOdd = Math.max(maxOdd,odd[str.charAt(i)-'a']);
                }
                System.out.println(n-maxEven-maxOdd);
            }
        }
    }
}

**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
1
a
2
ca
3
aab
5
ababa
6
acdada
9
ejibmyyju
6
bbccbc
6
abacba
5
bcbca
5
dcbdb
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Alternating String
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sakurako really loves alternating strings. She calls a string $s$ of lowercase Latin letters an alternating string if characters in the even positions are the same, if characters in the odd positions are the same, and the length of the string is even.

For example, the strings 'abab' and 'gg' are alternating, while the strings 'aba' and 'ggwp' are not.

As a good friend, you decided to gift such a string, but you couldn't find one. Luckily, you can perform two types of operations on the string:

  1. Choose an index $i$ and delete the $i$-th character from the string, which will reduce the length of the string by $1$. This type of operation can be performed no more than $1$ time;    2. Choose an index $i$ and replace $s_i$ with any other letter.

Since you are in a hurry, you need to determine the minimum number of operations required to make the string an alternating one.

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

The first line of each test case contains a single number $n$ ($1 \le n\le 2\cdot 10^5$)  — the length of the string.

The second line of each test case contains a string $s$, consisting of lowercase Latin letters.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output a single integer — the minimum number of operations required to turn the string $s$ into an alternating one.

Examples:
input:
10
1
a
2
ca
3
aab
5
ababa
6
acdada
9
ejibmyyju
6
bbccbc
6
abacba
5
bcbca
5
dcbdb
output:
1
0
1
1
2
6
2
3
1
1

Note: For the string ababa, you can delete the first character to get baba, which is an alternating string.

For the string acdada, you can change the first two characters to get dadada, which is an alternating string.

Buggy Code:
// Problem: E. Alternating String
// Contest: Codeforces - Codeforces Round 970 (Div. 3)
// URL: https://codeforces.com/contest/2008/problem/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// Author: Neil Chetty

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

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

  private static void solve() {
    // WRITE CODE HERE
    int n = in.nextInt();
    String s = in.nextLine();
    if (n % 2 == 0) {
      Data odd = new Data(), even = new Data();
      for (int i = 0; i < n; i++) {
        if (i % 2 == 0)
          even.add(s.charAt(i));
        else
          odd.add(s.charAt(i));
      }
      out.println(even.cost + odd.cost);
      return;
    }
    Data o = new Data(), e = new Data();
    Data[][] ok = new Data[n][2];
    for (int i = 0; i < n; i++) {
      ok[i][0] = Data.copy(o);
      ok[i][1] = Data.copy(e);
      if (i % 2 == 0)
        e.add(s.charAt(i));
      else
        o.add(s.charAt(i));
    }
    int result = Integer.MAX_VALUE;
    Data odd = new Data(), even = new Data();
    for (int i = n - 1; i >= 0; i--) {
      Data oo = Data.merge(ok[i][0], even), ee = Data.merge(ok[i][1], odd);
      result = Math.min(result, 1 + oo.cost + ee.cost);
      if (i % 2 == 0)
        even.add(s.charAt(i));
      else
        odd.add(s.charAt(i));
    }
    out.println(result);
  }

  // ALL PRE-WRITTEN HELPER FUNCTIONS
  static QuickReader in = new QuickReader();
  static PrintWriter out = new PrintWriter(System.out);
  private final static int MOD = (int) ((1e9) + 7);

  private static boolean[] primeFinder(int n) {
    boolean[] result = new boolean[n + 1];
    Arrays.fill(result, true);
    result[0] = false;
    result[1] = false;
    for (int i = 2; i * i <= n; i++) {
      if (!result[i])
        continue;
      for (int j = i * i; j <= n; j += i) {
        result[j] = false;
      }
    }
    return result;
  }

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

  private static long lcm(long a, long b) {
    return a * b / gcd(a, b);
  }

  private static long power(long a, long b) {
    long result = 1;
    while (b > 0) {
      if ((b & 1) == 1) {
        result = (result * a) % MOD;
      }
      b >>= 1;
      a = (a * a) % MOD;
    }
    return result;
  }

  private static long modInverse(long a) {
    return power(a, MOD - 2);
  }

  private static long nCr(int n, int r) {
    if (n < r)
      return 0;
    if (r == 0)
      return 1;
    long[] dp = factorial(n);
    return ((dp[n] * modInverse(dp[r]) % MOD) * (modInverse(dp[n - r]) % MOD)) % MOD;
  }

  private static long nCr(int n, int r, long[] dp) {
    if (n < r)
      return 0;
    if (r == 0)
      return 1;
    return ((dp[n] * modInverse(dp[r]) % MOD) * (modInverse(dp[n - r]) % MOD)) % MOD;
  }

  private static long nPr(int n, int r) {
    if (n < r)
      return 0;
    if (r == 0)
      return 1;
    long[] dp = factorial(n);
    return (dp[n] * modInverse(dp[n - r])) % MOD;
  }

  private static long nPr(int n, int r, long[] dp) {
    if (n < r)
      return 0;
    if (r == 0)
      return 1;
    return (dp[n] * modInverse(dp[n - r])) % MOD;
  }

  private static long[] factorial(int n) {
    long[] factorial = new long[n + 1];
    factorial[0] = 1;
    for (int i = 1; i <= n; i++) {
      factorial[i] = (factorial[i - 1] * i) % MOD;
    }
    return factorial;
  }

  private static void printArray(int[] a) {
    for (int i : a) out.print(i + " ");
    out.println();
  }

  private static void printlnArray(int[] a) {
    for (int i : a) out.println(i);
  }

  private static void printMatrix(int[][] a) {
    for (int[] i : a) printArray(i);
  }

  private static void sortArray(int[] a, int order) {
    ArrayList<Integer> l = new ArrayList<>();
    for (int i : a) l.add(i);
    if (order == -1)
      Collections.sort(l, Collections.reverseOrder());
    else
      Collections.sort(l);
    for (int i = 0; i < a.length; i++) a[i] = l.get(i);
  }

  private static class QuickReader {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");

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

    String next() {
      while (!st.hasMoreTokens()) 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());
    }

    int[] readIntArray(int n) {
      int[] a = new int[n];
      for (int i = 0; i < n; i++) a[i] = nextInt();
      return a;
    }

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

class Data {
  public int count, cost;
  public Character max;
  public Map<Character, Integer> map;

  public Data() {
    this.count = 0;
    this.cost = 0;
    this.max = null;
    this.map = new HashMap<>();
  }

  public void add(char c) {
    this.count++;
    this.map.put(c, map.getOrDefault(c, 0) + 1);
    if (this.max == null || this.map.get(c) >= this.map.get(max))
      this.max = c;
    this.cost = this.count - this.map.get(max);
  }

  public void add(char c, int value) {
    this.count += value;
    this.map.put(c, map.getOrDefault(c, 0) + value);
    if (this.max == null || this.map.get(c) >= this.map.get(max))
      this.max = c;
    this.cost = this.count - this.map.get(max);
  }

  public static Data merge(Data a, Data b) {
    Data res = Data.copy(a);
    for (char c : b.map.keySet()) res.add(c, b.map.get(c));
    return res;
  }

  public static Data copy(Data a) {
    Data res = new Data();
    for (char c : a.map.keySet()) res.add(c, a.map.get(c));
    return res;
  }
}

**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
abba
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

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
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define fr(i,n) for(int i= 0; i < (n); i++)
#define arrin(a,n)  for(int i=0;i<(n);i++) cin>>a[i];
#define printv(k) for(auto& i:k) cout<<i<<" "; cout<<endl;
#define printmp(k) for(auto& i:k) cout<<i.first<<" "<<i.second<<endl;
#define pb push_back
#define all(x)  x.begin(), x.end()
#define int long long
#define inf (1LL<<60)
#define endl '\n'
#define nl cout<<endl;
const int mod=1000000007;
const int mod2=998244353;
// priority_queue <pair <int , int> , vector <pair <int , int>> , greater <pair <int , int>>> pq;
string getBinary(int n) {
    bitset<64> b(n);
    return b.to_string();
}

void solve() {
    int n;
    cin>>n;
    string s,r;
    cin>>s>>r;
    stack<char> st;
    int idx = 0;
    st.push(s[0]);

    int i = 1;
    while(i < n){
        if(st.size() == 0) {
            st.push(s[i]);
            i++;
            continue;
        }
        char tp = st.top();
        if(tp == s[i]){
            st.push(s[i]);
            i++;
        }else {
            st.pop();
            char ch = r[idx++];
            while(st.size() && idx < n - 1){
                char tp = st.top();
                if(tp != ch){
                    st.pop();
                    ch = r[idx++];
                }else {
                    st.push(ch);
                    break;
                }
            }
            st.push(ch);
            i++;
        }
    }

    if(st.size() == 1){
        cout<<"YES"<<endl;
    }else {
        cout<<"NO"<<endl;
    }
}

signed main() {
    ios::sync_with_stdio(0), cin.tie(0);
    cout.tie(0);
    int T = 1;
    cin >> T;
    while (T--) {
        solve();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases that might cause the algorithm to fail. The primary issues in the code are:

1. **Handling of consecutive identical characters**: The code does not handle cases where consecutive identical characters in `s` are not correctly managed.
2. **Stack management**: The code might incorrectly manage the stack, especially when multiple replacements are required.

Given these points, let's construct a test case that can expose these issues:

### Test Case Construction:
1. **Length of `s` and `r`**: Choose a small length for `s` and `r` to ensure the test case is valid and within the constraints.
2. **Consecutive identical characters**: Ensure `s` has consecutive identical characters to test how the code handles them.
3. **Replacement sequence**: Ensure `r` has a sequence that might cause issues when replacing characters in `s`.

### Example Test Case:
```plaintext
1
5
11100
1010
```

### Explanation:
- **Input**:
  - `n = 5`
  - `s = "11100"`
  - `r = "1010"`

- **Expected Behavior**:
  - The code should correctly handle the consecutive identical characters in `s` and perform the replacements using `r`.
  - The sequence of operations should be:
    - First operation: Choose `k = 1` (11100 -> 1100)
    - Second operation: Choose `k = 1` (1100 -> 100)
    - Third operation: Choose `k = 1` (100 -> 10)
    - Fourth operation: Choose `k = 1` (10 -> 1)

- **Expected Output**:
  - The final string should be "1", and the output should be "YES".

### Output:
```plaintext
1
5
11100
1010
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

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
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

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

#define ull unsigned long long
#define ll long long
#define ui unsigned int
#define sortva(v) sort(v.begin(), v.end())
#define endl "\n"
#define fon for(int i=0; i<n; i++)
const ll MOD = 1e9 + 7;

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        string s;
        cin>>s;
        string r;
        cin>>r;
        int cnt1=0;
        int cnt0=0;
        for(int i=0; i<n; i++){
            if(s[i]=='1'){
                cnt1++;
            }
            if(s[i]=='0'){
                cnt0++;
            }
        }
        if(n==2){
            if(cnt1==1 || cnt0==1){
                cout<<"YES"<<endl;
            }else{
                cout<<"NO"<<endl;
            }
            continue;
        }
        bool flag = false;
        for(int i=0; i<n-1; i++){
            if(r[i]=='1'){
                cnt0--;
            }else{
                cnt1--;
            }
            if((cnt0==0 && cnt1>1)||(cnt1==0 && cnt0>1)){
                flag = true;
                break;

            }

        }
        if(!flag){
            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:
```plaintext
3
2
00
1
3
010
10
4
1110
000
```

### Explanation:
1. **Test Case 1:**
   - `n = 2`
   - `s = "00"`
   - `r = "1"`
   - **Expected Output:** `NO`
   - **Reason:** There are no two adjacent characters in `s` that are different, so you cannot perform the operation. Hence, you lose.

2. **Test Case 2:**
   - `n = 3`
   - `s = "010"`
   - `r = "10"`
   - **Expected Output:** `YES`
   - **Reason:** You can perform the following operations:
     - First operation: Choose `k = 1` (i.e., replace `s[1]s[2]` with `r[0]`), resulting in `s = "00"`.
     - Second operation: Choose `k = 1` (i.e., replace `s[1]s[2]` with `r[1]`), resulting in `s = "0"`.
     - You win the game.

3. **Test Case 3:**
   - `n = 4`
   - `s = "1110"`
   - `r = "000"`
   - **Expected Output:** `NO`
   - **Reason:** You can perform the following operations:
     - First operation: Choose `k = 2` (i.e., replace `s[2]s[3]` with `r[0]`), resulting in `s = "110"`.
     - Second operation: Choose `k = 1` (i.e., replace `s[1]s[2]` with `r[1]`), resulting in `s = "10"`.
     - Third operation: Choose `k = 1` (i.e., replace `s[1]s[2]` with `r[2]`), resulting in `s = "1"`.
     - However, after the first operation, you are left with `s = "110"`, and there are no two adjacent characters that are different, so you cannot perform the second operation. Hence, you lose.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

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
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

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

#define get(s)   \
    long long s; \
    cin >> s;

using namespace std;
int main()
{
    get(t);
    while (t--)
    {
        get(n);
        string sf;
        string rf;
        rf += '0';
        cin >> sf >> rf;
        vector<long long> s;
        vector<long long> r;
        for (long long j = 0; j < n; j++)
        {
            char c = sf[j];
            s.push_back(c - '0');
            char a = rf[j];
            r.push_back(a - '0');
        }
        int cow = 0;
        int OneCount = count(s.begin(), s.end(), 1);
        int ZeroCount = count(s.begin(), s.end(), 0);
        bool no = false;
        while ((OneCount + ZeroCount) != 1 && cow < n - 1)
        {
            if (OneCount == 0)
            {
                no = true;
                break;
            }
            if (ZeroCount == 0)
            {
                no = true;
                break;
            }
            if (OneCount >= 1 && ZeroCount >= 1)
            {
                if (r[cow] == 1)
                {
                    ZeroCount--;
                    cow++;
                }
                if (r[cow] == 0)
                {
                    OneCount--;
                    cow++;
                }
            }
        }
        if (no)
        {
            cout << "NO" << endl;
        }
        else
        {
            cout << "YES" << 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:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code has a few issues:

1. It does not correctly handle the transition of characters in the string `s`.
2. It does not correctly check the conditions for the existence of valid indices where `s[k] != s[k+1]`.

Let's consider a few edge cases:

1. **Case with a single transition point**: This can help test if the code correctly identifies and processes the transition.
2. **Case with multiple transition points**: This can help test if the code can handle multiple valid transitions.
3. **Case with no valid transition points**: This should result in a "NO" output, but the code might fail to handle it correctly.
4. **Case with alternating characters**: This can test if the code can handle a string with many valid transition points.

Given these considerations, a small-scale test input that can expose the bug might look like this:

```plaintext
3
4
1101
001
4
1010
101
4
1111
000
```

### Explanation:

1. **Test Case 1**:
   - `s = "1101"`, `r = "001"`
   - This string has multiple valid transition points (1-0 and 0-1).
   - The correct output should be "YES" because it is possible to perform the operations successfully.

2. **Test Case 2**:
   - `s = "1010"`, `r = "101"`
   - This string has multiple valid transition points (1-0, 0-1, 1-0).
   - The correct output should be "YES" because it is possible to perform the operations successfully.

3. **Test Case 3**:
   - `s = "1111"`, `r = "000"`
   - This string has no valid transition points (all characters are the same).
   - The correct output should be "NO" because it is impossible to perform any operation.

This test input should expose the bug in the provided code by testing different scenarios where the code might fail.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Replacement
time_limit_ms: 1000
memory_limit_mb: 256
Description: You have a binary string$^{\text{∗}}$ $s$ of length $n$, and Iris gives you another binary string $r$ of length $n-1$.

Iris is going to play a game with you. During the game, you will perform $n-1$ operations on $s$. In the $i$-th operation ($1 \le i \le n-1$):

  * First, you choose an index $k$ such that $1\le k\le |s| - 1$ and $s_{k} \neq s_{k+1}$. If it is impossible to choose such an index, you lose;    * Then, you replace $s_ks_{k+1}$ with $r_i$. Note that this decreases the length of $s$ by $1$.

If all the $n-1$ operations are performed successfully, you win.

Determine whether it is possible for you to win this game.

$^{\text{∗}}$A binary string is a string where each character is either $\mathtt{0}$ or $\mathtt{1}$.

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

The first line of each test case contains a single integer $n$ ($2\le n\le 10^5$) — the length of $s$.

The second line contains the binary string $s$ of length $n$ ($s_i=\mathtt{0}$ or $\mathtt{1}$).

The third line contains the binary string $r$ of length $n-1$ ($r_i=\mathtt{0}$ or $\mathtt{1}$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$.

Output: For each test case, print "YES" (without quotes) if you can win the game, and "NO" (without quotes) otherwise.

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
11
0
2
01
1
4
1101
001
6
111110
10000
6
010010
11010
8
10010010
0010010
output:
NO
YES
YES
NO
YES
NO

Note: In the first test case, you cannot perform the first operation. Thus, you lose the game.

In the second test case, you can choose $k=1$ in the only operation, and after that, $s$ becomes equal to $\mathtt{1}$. Thus, you win the game.

In the third test case, you can perform the following operations: $\mathtt{1}\underline{\mathtt{10}}\mathtt{1}\xrightarrow{r_1=\mathtt{0}} \mathtt{1}\underline{\mathtt{01}} \xrightarrow{r_2=\mathtt{0}} \underline{\mathtt{10}} \xrightarrow{r_3=\mathtt{1}} \mathtt{1}$.

Buggy Code:
def find_max_consecutive(s):
    max_0 = 0
    max_1 = 0
    current_0 = 0
    current_1 = 0

    for ch in s:
        if ch == '0':
            current_0 += 1
            max_1 = max(max_1, current_1)  # Update max_1 before resetting current_1
            current_1 = 0  # Reset current_1
        else:
            current_1 += 1
            max_0 = max(max_0, current_0)  # Update max_0 before resetting current_0
            current_0 = 0  # Reset current_0

    # Final update in case the string ends with a sequence
    max_0 = max(max_0, current_0)
    max_1 = max(max_1, current_1)

    return max_0, max_1

def main():
    t = int(input())  # Read number of test cases

    results = []
    for _ in range(t):
        n = int(input())  # Read the length of s
        s = input()       # Read the string s
        r = input()       # Read the string r

        max_0_s, max_1_s = find_max_consecutive(s)  # Find max consecutive in s
        max_0_r, max_1_r = find_max_consecutive(r)  # Find max consecutive in r

        # Check if max_0 in s equals max_1 in r and max_1 in s equals max_0 in r
        if (max_0_s == max_1_r and max_1_s == max_0_r) or (n == 2 and (s == "01" or s == "10")):
            results.append("YES")
        else:
            results.append("NO")

    # Output all results
    for result in results:
        print(result)

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases where the conditions in the `main` function might fail. The key conditions to check are:

1. The maximum consecutive zeros in `s` should equal the maximum consecutive ones in `r`.
2. The maximum consecutive ones in `s` should equal the maximum consecutive zeros in `r`.
3. Special handling for strings of length 2.

The buggy code checks these conditions, but it might fail in cases where the sequences of zeros and ones in `s` and `r` are not balanced in the way the code expects. Let's construct a test input that can expose this:

1. **Test Case 1: Simple Case with Length 2**
   - `s = "01"` and `r = "1"`. This should return "YES".

2. **Test Case 2: Simple Case with Length 2**
   - `s = "10"` and `r = "0"`. This should return "YES".

3. **Test Case 3: Longer String with Balanced Consecutive Zeros and Ones**
   - `s = "11010011"` and `r = "001101"`. This should return "NO" because the maximum consecutive zeros in `s` (2) do not equal the maximum consecutive ones in `r` (2).

4. **Test Case 4: Longer String with Unbalanced Consecutive Zeros and Ones**
   - `s = "111000111000"` and `r = "111000"`. This should return "NO" because the maximum consecutive zeros in `s` (3) do not equal the maximum consecutive ones in `r` (3).

5. **Test Case 5: String with No Consecutive Zeros or Ones**
   - `s = "010101"` and `r = "10101"`. This should return "NO" because there are no consecutive zeros or ones in `s` or `r`.

6. **Test Case 6: String with All Zeros or All Ones**
   - `s = "000000"` and `r = "00000"`. This should return "NO" because there are no operations that can be performed.

Here is the test input:

```plaintext
6
2
01
1
2
10
0
8
11010011
001101
12
111000111000
111000
6
010101
10101
6
000000
00000
```

This test input includes a variety of cases to ensure that the buggy code is thoroughly tested.