**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

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

using namespace std;

#define Code ios_base::sync_with_stdio(false);
#define By cin.tie(NULL);
#define PRAYAG cout.tie(NULL);

#define ll long long
#define lld long double
#define ull unsigned long long

const lld pi = 3.141592653589793238;
const ll INF = 1e18;
const ll mod = 1e9 + 7;
// const ll mod=998244353;

typedef unordered_map<ll, ll> umll;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef vector<pll> vpll;
typedef vector<string> vs;
typedef unordered_map<ll, ll> umll;
typedef map<ll, ll> mll;
#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define vi vector<int>
#define vs vector<string>
#define vvi vector<vector<int>>
#define vvs vector<vector<string>>
#define vpi vector<pair<int, int>>
#define vv(T) vector<vector<T>>
#define maxhp(T) priority_queue<T>
#define minhp(T) priority_queue<T, vector<T>, greater<T>>
#define mem(n, i) memset(n, i, sizeof n)
#define p(A, B) pair<A, B>

ll gcd(ll a, ll b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
ll lcm(ll a, ll b) { return (a / gcd(a, b) * b); }
ll moduloMultiplication(ll a, ll b, ll mod)
{
    ll res = 0;
    a %= mod;
    while (b)
    {
        if (b & 1)
            res = (res + a) % mod;
        b >>= 1;
    }
    return res;
}
ll powermod(ll x, ll y, ll p)
{
    ll res = 1;
    x = x % p;
    if (x == 0)
        return 0;
    while (y > 0)
    {
        if (y & 1)
            res = (res * x) % p;
        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}
ll mod_mult(ll a, ll b, ll m) { return (a % m * b % m) % m; }
ll mod_mul_inverse(ll a, ll b) { return powermod(a, b - 2, b); }
ll mod_div(ll a, ll b, ll m) { return mod_mult(a, mod_mul_inverse(b, m), m); }
ll MSBPosition(ll N)
{
    ll msb_p = -1;
    while (N)
    {
        N = N >> 1;
        msb_p++;
    }
    return msb_p;
}

bool sorta(const pair<int, int> &a, const pair<int, int> &b) { return (a.second < b.second); }
bool sortd(const pair<int, int> &a, const pair<int, int> &b) { return (a.second > b.second); }
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) { return (a.second < b.second); }

bool isPrime(ll n)
{
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
    if (n % 2 == 0 || n % 3 == 0)
        return false;
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
    return true;
}
bool isPowerOfTwo(int n)
{
    if (n == 0)
        return false;
    return (ceil(log2(n)) == floor(log2(n)));
}
bool isPerfectSquare(ll x)
{
    if (x >= 0)
    {
        ll sr = sqrt(x);
        return (sr * sr == x);
    }
    return false;
}
ll fastprime(ll a)
{
    if (a != 2 && a % 2 == 0 || a < 2)
        return 0;
    for (ll i = 3; i * i <= a; i += 2)
        if (a % i == 0)
            return 0;
    return 1;
}

#define TO_LOWERCASE(s) transform((s).begin(), (s).end(), (s).begin(), [](char c) { return tolower(c); })
#define SAR(s)                                                     \
    do                                                             \
    {                                                              \
        std::sort((s).begin(), (s).end());                         \
        (s).erase(std::unique((s).begin(), (s).end()), (s).end()); \
    } while (0)
#define FSUB(x, s) ((x).find(s) != std::string::npos ? (x).find(s) : -1)
//     sort(ord.begin(), ord.end(), [&](auto x, auto y) { return p[x] < p[y]; });
//     string string=to_string(num);
//     ll int num=stoi(string);
int ask(int l, int r)
{
    cout << "? " << l << ' ' << r << endl;
    fflush(stdout);
    int ans;
    cin >> ans;
    return ans;
}

#define py cout << "YES" << "\n"
#define pn cout << "NO" << "\n"
#define pm cout << "-1" << endl;

#define ff first
#define ss second
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define all(v) v.begin(), v.end()
#define allr(v) v.rbegin(), v.rend()
#define rv(v) v.end(), v.begin()

#define ps(x, y) fixed << setprecision(y) << x;
void SieveOfEratosthenes(ll n, vll &v)
{
    bool prime[n + 1];
    memset(prime, true, sizeof(prime));
    for (int p = 2; p * p <= n; p++)
    {
        if (prime[p] == true)
        {
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
    for (int p = 2; p <= n; p++)
        if (prime[p])
            v.eb(p);
}
map<ll, set<ll>> m;
void dfs(ll root)
{
    if (m[root].size() == 0)
    {
        return;
    }
    for (auto it : m[root])
    {
        m[it].erase(root);
        dfs(it);
    }
}

#define o(x) cout << (x) << " ";
#define o2(x, y) cout << (x) << " " << (y) << " ";
#define o3(x, y, z) cout << (x) << " " << (y) << " " << (z) << " ";

#define oe(x) cout << (x) << endl;
#define oe2(x, y) cout << (x) << " " << (y) << endl;
#define oe3(x, y, z) cout << (x) << " " << (y) << " " << (z) << endl;

#define i1(A, a) \
    A a;         \
    cin >> a;
#define i2(A, a, b) \
    A a, b;         \
    cin >> a >> b;
#define i3(A, a, b, c) \
    A a, b, c;         \
    cin >> a >> b >> c;
#define i4(A, a, b, c, d) \
    A a, b, c, d;         \
    cin >> a >> b >> c >> d;
#define i5(A, a, b, c, d, e) \
    A a, b, c, d, e;         \
    cin >> a >> b >> c >> d >> e;
#define i6(A, a, b, c, d, e, f) \
    A a, b, c, d, e, f;         \
    cin >> a >> b >> c >> d >> e >> f;

#define inarr(arr, n)          \
    for (ll i = 0; i < n; i++) \
    cin >> arr[i]

#define fl(i, n) for (ll i = 0; i < n; i++)
#define Fl(i, k, n) for (ll i = k; i < n; i++)
#define bl(i, n) for (ll i = n; i >= 0; i--)
#define Bl(i, k, n) for (ll i = n; i >= k; i--)

// rotate(v.begin(),v.begin()+v.size()-r,v.end()); for rotating vector r times right
// rotate(v.begin(),v.begin()+r,v.end()); for rotating vector r times left

const ll N = 1e6 + 5;
void solve()
{

    ll int n, k;
    cin >> n >> k;

    vll f(N);
    f[0] = 1;
    f[1] = 1;
    if (k == 1)
    {
        cout << n << '\n';
        return;
    }
    for(ll int i=2;;i++)
    {
        if (i >= 2)
            f[i] = (f[i - 1] + f[i - 2]) % k;
        if (f[i] % k == 0)
        {
            n %= mod;
            oe(n * (i + 1) % mod)

                return;
        }
    }
}

int main()
{

    Code By PRAYAG

        int tt = 1;
    cin >> tt;
    while (tt--)
    {
        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 buggy code, let's analyze the code and identify potential issues:

1. **Fibonacci Calculation Modulo k**: The code calculates the Fibonacci sequence modulo `k` to find the index of the Fibonacci number that is divisible by `k`. This is done in a loop that continues indefinitely until a Fibonacci number divisible by `k` is found.

2. **Handling Large `n`**: The code directly multiplies `n` by the index `i + 1` and takes the result modulo `10^9 + 7`. This is a potential issue because `n` can be as large as `10^18`, and the multiplication can cause overflow or incorrect results.

3. **Edge Cases**: The code does not handle edge cases well, such as when `k` is 1 or when `n` is very large.

Given these points, we can create a test case that maximizes the failure probability by using a large `n` and a `k` that will make the Fibonacci sequence take a long time to find a number divisible by `k`.

### Chosen Test Case
- **`n`**: A large number, e.g., `10^18 - 1` (just under the maximum value).
- **`k`**: A value that is not a common divisor of early Fibonacci numbers, e.g., `1377` (from the example).

This test case will:
- Ensure that the Fibonacci sequence modulo `k` takes a long time to find a number divisible by `k`.
- Test the handling of large `n` and the correctness of the multiplication and modulo operations.

### Test Input
```plaintext
1
999999999999999999 1377
```

This test input is small and valid, and it should expose the issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define all(x) x.begin(),x.end()
typedef long long ll;
#ifdef GYAN
    #include"ratan/debug.h"
#else
    #define dbg(...)
    #define dbl(...)
#endif
template<typename... Args> void in(Args&... args){((cin >> args), ...);}
// ---------------------------------------------------------------------------
const ll mod= 1e9+7;
const ll mx= 1e6;
vector<ll> fib(mx+2,1);
void solve(int tt){
    dbg("TestCase: "+to_string(tt));
    ll n,k;
    cin>>n>>k;
    if(k==1){
        cout<<n<<'\n';
        return;
    }
    for(ll i=3;i<=k+1;i++){
        fib[i] = (fib[i-1]+fib[i-2])%k;
    }
    ll res=-1;
    n%=mod;
    ll i=1;
    while(true){
        // dbg(fib[i]%k);
        if(fib[i]==0){
            res=i;
            break;
        }
        i+=1;
    }
    cout<<(n*res)%mod<<'\n';
}
signed main(){
    ios_base::sync_with_stdio(0);cin.tie(0);
    int t;
    cin>>t;
    int T=t;
    while(t--) solve(T-t);
    return 0;
}

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

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

Think step by step.

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

1. **Edge Cases**: Values of `n` and `k` that are at the boundary conditions.
2. **Large Values**: Values of `n` and `k` that are large to test the performance and correctness of the algorithm.
3. **Special Cases**: Values of `n` and `k` that might cause unexpected behavior, such as `k=1` or `n` being very large.

The bug in the code is that it precomputes the Fibonacci sequence modulo `k` up to `k+1`, which is not sufficient for large values of `n`. Additionally, the code does not handle the case where `n` is very large correctly.

Let's create a test input that includes:
- A test case where `k` is 1.
- A test case where `n` is a large number and `k` is a small number.
- A test case where both `n` and `k` are large.

Here is the test input:

```plaintext
3
1 1
1000000000000 2
999999999999999999 99999
```

### Explanation:
1. **Test Case 1**: `n = 1`, `k = 1`
   - This is a simple case where `k = 1`. The expected output is `1` because every Fibonacci number is divisible by `1`.

2. **Test Case 2**: `n = 1000000000000`, `k = 2`
   - This is a large value of `n` with a small `k`. The expected output is the 3rd Fibonacci number that is divisible by 2, which is `34`, and the index is `9`. The result should be `9` modulo `10^9+7`.

3. **Test Case 3**: `n = 999999999999999999`, `k = 99999`
   - This is a very large value of `n` with a large `k`. The expected output should be the index of the 999999999999999999th Fibonacci number that is divisible by 99999, modulo `10^9+7`.

This test input should help expose the bugs in the provided code, particularly the insufficient precomputation and the incorrect handling of large values of `n`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
from collections import defaultdict
import math

MOD = 10**9 + 7
MAX_FIB_LENGTH = 10001
fibs = [1, 1]

def fib():
    for i in range(2, MAX_FIB_LENGTH):
        fibs.append(fibs[i - 1] + fibs[i - 2])

fib()

def solve():
    n, k = map(int, input().split())

    for i in range(min(n * k, MAX_FIB_LENGTH)):
        if fibs[i] % k == 0:
            print((n * (i + 1))%MOD)
            return

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

if __name__ == '__main__':
    main()

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the key points to consider:

1. **Large `n` and `k` values**: The code uses a fixed maximum length for the Fibonacci sequence (`MAX_FIB_LENGTH = 10001`). If `n` and `k` are large, the code might not find the required Fibonacci number within this limit, leading to incorrect results.

2. **Fibonacci sequence modulo `k`**: The code checks if the Fibonacci number is divisible by `k` and then prints the index. However, it only checks up to `MAX_FIB_LENGTH`, which might not be sufficient for large `n` and `k`.

3. **Efficiency and time complexity**: The code might run into performance issues for large `n` and `k` due to the fixed limit on the Fibonacci sequence.

Given these considerations, a good test input would be one where the required Fibonacci number is beyond the `MAX_FIB_LENGTH` limit, but still within the problem's constraints.

### Test Input
```plaintext
1
1000000000000 1377
```

### Explanation
- **`n = 1000000000000`**: This is a very large value for `n`, which will require a very large Fibonacci number.
- **`k = 1377`**: This is a relatively large value for `k`, making it more likely that the required Fibonacci number will be beyond the `MAX_FIB_LENGTH` limit.

This test input will likely expose the bug in the code because the code will not find the required Fibonacci number within the first 10001 terms, leading to an incorrect result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
import sys
I = input = lambda: sys.stdin.readline().rstrip()
II, LI = lambda: int(I()), lambda: list(map(int, I().split()))
oo, M = 1 << 63, 1000000007 # 998244353
def print(*x, sep=' '): sys.stdout.write(sep.join(map(str, x))+'\n')

def solve():
    n, k = LI()
    F = [1, 1]
    if k == 1:
        return print(n % M)
    for _ in range(k):
        f = (F[-1] + F[-2]) % k
        F.append(f)
        if f == 0:
            break
    ans = 0
    ln = len(F)
    done = False
    while not done and ln >= 2:
        b = F[ln - 2]
        done = True
        ans += ln; ans %= M; n -= 1
        if not n: return print(ans)
        for i in range(ln - 1):
            if not F[i] * b % k:
                ln = i + 1
                done = False
                break
    if n > 0:
        ans += ln * n % M
    print(ans)

for _ in range(II()):
    solve()

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can cause the code to fail. Here are the key points to consider:

1. **Edge Cases for \( n \) and \( k \)**:
   - \( n = 1 \) or \( n = 2 \)
   - \( k = 1 \)
   - \( k \) that causes the Fibonacci sequence to have a very large period before hitting a multiple of \( k \).

2. **Large Values**:
   - \( n \) and \( k \) are at the upper bounds of their constraints.

3. **Special Cases**:
   - \( k \) is a prime number.
   - \( k \) is a composite number with specific properties (e.g., \( k = 2 \)).

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

```plaintext
5
1 1
2 1
3 2
1000000000000 1
1000000000000 1377
```

### Explanation:
1. **Test Case 1: \( n = 1 \), \( k = 1 \)**
   - This is a basic edge case where the first Fibonacci number is divisible by 1.
   - Expected output: \( 1 \).

2. **Test Case 2: \( n = 2 \), \( k = 1 \)**
   - This is another basic edge case where the second Fibonacci number is divisible by 1.
   - Expected output: \( 2 \).

3. **Test Case 3: \( n = 3 \), \( k = 2 \)**
   - This is a simple case where the third Fibonacci number that is divisible by 2 is 34.
   - Expected output: \( 9 \).

4. **Test Case 4: \( n = 1000000000000 \), \( k = 1 \)**
   - This is a large \( n \) with \( k = 1 \).
   - Expected output: \( 1000000000000 \mod 1000000007 \).

5. **Test Case 5: \( n = 1000000000000 \), \( k = 1377 \)**
   - This is a large \( n \) with a specific \( k \) that can cause the Fibonacci sequence to have a large period.
   - Expected output: \( 999244007 \).

This test input should be sufficient to expose any bugs in the provided code, especially in handling large values and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
mod=10**9+7
for t in range(int(input())):
    n,k=map(int, input().split())
    if k==1:
        print(n%mod)
        continue
    f=[0]*(k+1)
    f[0]=1
    f[1]=1
    y=0
    for i in range(2,k+1):
        f[i]=f[i-1]+f[i-2]
        f[i]%=k
        if f[i]==0:
            y=i+1
            break
    ans=y*(n%mod)
    ans%=mod
    print(ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
// Source: https://usaco.guide/general/io

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

public class Main {
	public static long binexp(long a, long b, long MOD){
		long mult = 1;
		while(b >= 1){
			if(b%2 == 1){
				mult = (mult * a)%MOD;
			}
			b /= 2;
			a = (a * a)%MOD;

		}
		return mult;
	}
	public static long gcd(long a, long b){
		return (a < b) ? gcd(b,a) : (a%b == 0 ? b : gcd(a%b,b));
	}
	public static long lcm(long a, long b){
		return a/gcd(a,b) * b;
	}
	public static long modinv(long a, long MOD){
		return binexp(a, MOD - 2, MOD);
	}
	public static boolean isPrime(int number){
		if(number <= 1) return false;
		for(int p = 2; p <= Math.sqrt(number); p++){
			if(number%p == 0) return false;
		}
		return true;

	}
	public static Map<Integer,Integer> primeFact(int num){
		Map<Integer,Integer> map = new HashMap<Integer,Integer>();
		List<Integer> factors = new ArrayList<Integer>();
		for(int i = 1; i <= Math.sqrt(num); i++){
			if(num%i == 0){
				if(isPrime(i)) factors.add(i);
				else if(isPrime(num/i)) factors.add(num/i);
			}

		}
		//System.out.println(factors);
		for(Integer fact : factors){
			while(num%fact == 0){
				num /= fact;
				map.put(fact, map.getOrDefault(fact,0) + 1);
			}
		}
		//System.out.println(map);
		return map;
	}
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		long MOD = 1000000007;
		while(T-- > 0){
			long N = sc.nextLong();
			int K = sc.nextInt();
			long ans = 1;
			Map<Integer,Integer> f = primeFact(K);
			for(Integer factor : f.keySet()){
				//System.out.println(factor + " " + f.get(factor));
				ans = lcm(ans, ((factor + 1) * binexp(factor, f.get(factor) - 1, MOD))%MOD);
			}

			System.out.println((ans * N)%MOD);
		}
	}
}

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

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

Think step by step.

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

1. **Edge Cases**: The input should test the limits of the given constraints, such as the maximum values for \( n \) and \( k \).
2. **Special Values**: Values that might cause the code to fail due to specific conditions, such as large prime numbers or composite numbers with specific factorizations.
3. **Behavior on Large Inputs**: Ensure the input is large enough to stress the algorithm but small enough to be a valid test case within the constraints.

Given the constraints:
- \( 1 \le t \le 10^4 \)
- \( 1 \le n \le 10^{18} \)
- \( 1 \le k \le 10^5 \)

We will create a test input with a small number of test cases that include:
- A large value of \( n \)
- A value of \( k \) that is a large prime number
- A value of \( k \) that is a composite number with specific factorization

Here is a challenging test input:

```plaintext
3
1000000000000 999983
1000000000000 100000
1000000000000 1
```

Explanation:
- The first test case uses \( n = 10^{12} \) and \( k = 999983 \), which is a large prime number.
- The second test case uses \( n = 10^{12} \) and \( k = 100000 \), which is a composite number.
- The third test case uses \( n = 10^{12} \) and \( k = 1 \), which is a special case where the answer should be \( n \) itself.

This input should help expose any issues with the handling of large numbers and the prime factorization in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
import java.util.*;
public class fibon
{
    public static void main(String args[])
    {
        Scanner sc = new Scanner(System.in);
        int cases = sc.nextInt();

        while(cases-->0)
        {
            long n = sc.nextLong();
            long k = sc.nextLong();
            long arr[] = new long[200000];
            arr[0]=1%k;
            arr[1]=1%k;
            for(int i=2;i<2*k;i++)
            {
                arr[i]=((arr[i-1]%k)+(arr[i-2]%k))%k;
            }
            int i=0;
            inner:
            while(true)
            {
                if(arr[i]==0)
                break inner;

                i++;
            }
            i+=1;
            long sum=0;
            long mod=1000000007;
            for(int j=1;j<=i;j++)
            {
                sum=((sum%mod)+(n%mod))%mod;
            }
            System.out.println(sum);
        }
    }
}

**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, let's analyze the code and identify potential issues:

1. **Fibonacci Sequence Calculation**:
   - The code uses a fixed array size of 200,000 to store Fibonacci numbers modulo `k`.
   - The loop for generating Fibonacci numbers runs only up to `2*k`.

2. **Finding the First Zero**:
   - The code searches for the first zero in the array, which indicates a Fibonacci number divisible by `k`.
   - It then uses this index to compute the final result.

3. **Sum Calculation**:
   - The code calculates the sum of `n` repeated `i` times, where `i` is the index of the first zero in the array.

### Potential Issues:
- **Array Size**: The array size of 200,000 might not be sufficient for large values of `k`.
- **Loop Limit**: The loop limit of `2*k` might not be sufficient to find the first zero for large `k`.
- **Sum Calculation**: The sum calculation might overflow or produce incorrect results due to the way it is computed.

### Challenging Test Input:
To expose these issues, we can use a test case with a large value of `k` and a moderate value of `n` to ensure that the Fibonacci sequence modulo `k` does not fit within the array size or loop limit.

Here is a small-scale test input that should challenge the code:

```plaintext
1
100000 100000
```

This test case uses:
- `t = 1` (one test case)
- `n = 100000` (a large value of `n`)
- `k = 100000` (a large value of `k`)

This input will force the code to generate a large number of Fibonacci numbers modulo `100000`, which is likely to exceed the array size and loop limit, thus exposing the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
from functools import cache
t = int(input())
fib = []
@cache
def f(num):
    if num == 0 or num == 1:
        return 1
    return f(num-1)+f(num-2)
for i in range(0, 1500):
    fib.append(f(i))
MOD = 10**9+7
#print(fib)
for _ in range(t):
    n, k = map(int, input().split())
    first_ind = 0
    for i in range(len(fib)):
        if fib[i]%k == 0:
            first_ind = i+1
            break
    #first ind is just the first fibonacci number divisible by k
    #clear pattern to notice here is the distance between subsequent fibonacci numbers divisible by k
    #is equal to its first index
    print((first_ind + first_ind * (n-1))%MOD)

**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
1000000000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
MOD = 1e9+7

def solve():
	n,k = map(int,input().split())

	last = [1,0]
	i = 0
	while i == 0 or last[-1] != 0:
		x = sum(last) % k
		last = last[1:] + [x]
		i += 1
	n %= MOD
	ans = n * i
	ans %= MOD
	print(int(ans))

for _ in range(int(input())):
	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 buggy code, let's analyze the code and identify potential issues:

1. **Fibonacci Sequence Calculation**: The code uses a loop to find the period of the Fibonacci sequence modulo `k`. This is a valid approach, but it needs to be done correctly.
2. **Modulo Operation**: The code takes `n % MOD` before computing the final answer, which is incorrect. The period of the Fibonacci sequence modulo `k` should be used to find the correct index, not `n % MOD`.
3. **Multiplication and Final Modulo**: The code multiplies `n` by the period `i` and then takes the result modulo `MOD`. This is also incorrect because the period should be used to find the correct index in the sequence.

To expose these issues, we can use the following test cases:

1. **Small `n` and `k`**: To ensure the basic logic is correct.
2. **Large `n` and small `k`**: To test the handling of large numbers.
3. **Edge case for `k`**: To test the handling of specific values of `k`.

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

```plaintext
3
3 2
1000000000000 1
1000000000000 1377
```

This test input includes:
1. A small `n` and `k` to test the basic logic.
2. A large `n` with `k = 1` to test the handling of large numbers and the modulo operation.
3. A large `n` with a specific `k` to test the handling of specific values and the period calculation.

This should help expose the bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int t;
    long long div=1e9+7;
    cin>>t;
    while(t--){
        long long n,k;
        cin>>n>>k;
        if(k==1){
            cout<<n<<endl;
            continue;
        }
        long long x=2;
        long long prv1=1, prv2=1;
        long long ct=3;
        while((x%k)!=0){
            ++ct;
            prv2=prv1%k;
            prv1=x%k;
            x=prv1+prv2;
            x=x%k;
        }

        long long ans = ((n % div)*(ct % div) )% div;

        cout<<ans<<endl;
    }

    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios where the code might fail. The code calculates the index of the nth Fibonacci number that is divisible by k, and it uses a while loop to find the first Fibonacci number divisible by k. However, the code might have issues with large values of n and k, or with specific edge cases.

Here are some potential issues to consider:
1. **Large values of n**: The code might run into performance issues or incorrect results for very large n.
2. **Small values of k**: The code might not handle small values of k correctly, especially when k is 1.
3. **Specific Fibonacci sequences**: The code might fail for specific sequences where the Fibonacci numbers have unique properties.

Given these considerations, let's generate a test input that includes edge cases and potentially challenging scenarios:

1. **k = 1**: This should always return n, as every Fibonacci number is divisible by 1.
2. **n = 1 or n = 2**: These are the base cases for Fibonacci numbers.
3. **Large n and small k**: This can help check the performance and correctness for large n.
4. **k = 2**: This is a simple case where the Fibonacci sequence has a clear pattern (every third Fibonacci number is divisible by 2).

Here is a small-scale test input that includes these scenarios:

```plaintext
4
1 1
2 1
3 2
1000000000000 2
```

This input includes:
- A test case with n = 1 and k = 1.
- A test case with n = 2 and k = 1.
- A test case with n = 3 and k = 2.
- A test case with a large n and k = 2.

These test cases should help expose any issues in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

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

#define int long long
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define PB push_back
#define FS first
#define SD second
#define ary(k) array<int,k>
template<class A,class B> void cmx(A& x, B y) {x=max<A>(x,y);}
template<class A,class B> void cmn(A& x, B y) {x=min<A>(x,y);}
typedef pair<int, int> pii;
typedef vector<int> vi;

int mod = 1e9 + 7;

signed main() {
    cin.tie(0)->sync_with_stdio(0); cin.exceptions(cin.failbit);

    int t;
    cin >> t;
    while (t--) {
        int n, k;
        cin >> n >> k;

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

        // first is how many steps to get to next 0, second is what next number is
        pii nums = {1, 1};
        int curIdx = 1, numZeros = 0;
        vector<int> zeroOffsets = {curIdx};
        while (k > 1) {
            while (nums.first != 0) {
                int total = (nums.first + nums.second) % k;
                nums.first = nums.second;
                nums.second = total;
                curIdx++;
                // cout << nums.first << " " << nums.second << endl;
            }
            numZeros++;
            zeroOffsets.push_back(curIdx - zeroOffsets.back());
            int m_gcd = __gcd(nums.second, k);
            if (m_gcd == 1) {
                break;
            } else {
                k /= m_gcd;
                nums.second /= m_gcd;
            }
        }
        // cout << curIdx << " " << numZeros << endl;
        // in first curIdx nums, there are numZeros 0's
        // n * curIdx / numZeros
        int result = (n % mod);
        numZeros %= mod;
        int m_div = result / (numZeros);
        int m_mod = result % numZeros;
        result = ((m_div * curIdx - 1) % mod + zeroOffsets[m_mod]) % mod;
        cout << result << 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 given code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. The code seems to handle the general case of finding the index of the nth Fibonacci number divisible by k, but there are a few potential issues to consider:

1. **Small values of k**: The code might not handle small values of k correctly, especially when k = 1.
2. **Large values of n**: The code might have issues with very large values of n due to the modulo operation and the way it calculates the result.
3. **Special values of k**: Values of k that are prime or have special properties might cause the code to fail.
4. **Boundary conditions**: Edge cases like the minimum and maximum values of n and k should be tested.

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

```plaintext
4
1 1
2 1
1000000000000 1
1000000000000 100000
```

Explanation:
1. The first test case (`1 1`) checks the smallest possible value of n and k.
2. The second test case (`2 1`) checks another small value of n and k.
3. The third test case (`1000000000000 1`) checks a very large value of n with k = 1.
4. The fourth test case (`1000000000000 100000`) checks a very large value of n with a large value of k.

This test input should help identify any issues in the code, especially in handling edge cases and large values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.*;

public class f {
    static FastReader fr = new FastReader();
    static FastWriter fw = new FastWriter();
    static long dp[];
    public static void solve(){
        long n = fr.nextLong();
        int k = fr.nextInt();
        int mod = (int)1e9+7;
        if(k==1){
            fw.println(n);
            return;
        }
        long a =1,b=1;
        int cnt = 2;
        while(true){
           long c = b+a;
           cnt++;
           if(c%k==0) break;
           a = b%k; b = c%k;
        }
        //System.out.println(cnt);
        long ans = ((n%mod) * (cnt%mod))%mod;
        fw.println(ans);
    }
    public static void main(String[] args) throws NumberFormatException, IOException{
        //long start = System.nanoTime();
        int t = fr.nextInt();
        int N = 10000000;
        dp = new long[N];
        dp[1] = 1; dp[2] = 1;
        for(int i=3;i<N;i++) dp[i] = dp[i-1] + dp[i-2];
        //System.out.println(solve());
        //System.out.println(solve()?"YES":"NO");
        while(t-->0){
            solve();
            //System.out.println(solve());
            //System.out.println(solve()?"YES":"NO");
        }
        //System.out.println(((long)System.nanoTime()-start)/1000000);
    }
}

class FastReader{
    BufferedReader br;
    StringTokenizer st;
    public FastReader(){
        br = new BufferedReader(new InputStreamReader(System.in));
    }
    String next(){
        while (st == null || !st.hasMoreElements()){
            try{
                st = new StringTokenizer(br.readLine());
            }catch (IOException e){
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
    int nextInt(){
        return Integer.parseInt(next());
    }
    long nextLong(){
        return Long.parseLong(next());
    }
    double nextDouble(){
        return Double.parseDouble(next());
    }
    String nextLine(){
        String str = "";
        try{
            str = br.readLine();
        }
        catch (IOException e){
            e.printStackTrace();
        }
        return str;
    }
    int[] nextIntArray() {
        String line = nextLine();
        String[] strNums = line.split("\\s+");
        int[] arr = new int[strNums.length];
        for (int i = 0; i < strNums.length; i++) {
            arr[i] = Integer.parseInt(strNums[i]);
        }
        return arr;
    }
    long[] nextLongArray() {
        String line = nextLine();
        String[] strNums = line.split("\\s+");
        long[] arr = new long[strNums.length];
        for (int i = 0; i < strNums.length; i++) {
            arr[i] = Long.parseLong(strNums[i]);
        }
        return arr;
    }
    char[] nextCharArray() {
        String line = nextLine();
        String[] strNums = line.split("\\s+");
        char[] arr = new char[strNums.length];
        for (int i = 0; i < strNums.length; i++) {
            arr[i] = strNums[i].charAt(0);
        }
        return arr;
    }
    char[] StringtoChar() {
        String line = nextLine();
        return line.toCharArray();
    }
}

class FastWriter {
    private final PrintWriter writer;

    public FastWriter() {
        writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), true);
    }

    public void print(Object object) {
        writer.print(object);
    }

    public void println(Object object) {
        writer.println(object);
    }

    public void flush() {
        writer.flush();
    }
}

**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 test input that exposes the bug in the provided code, we need to consider the following aspects:

1. **Large Values of \( n \)**: The code might fail when \( n \) is very large, especially when \( n \) is around \( 10^{18} \). The algorithm should handle such large values efficiently.
2. **Small Values of \( k \)**: The code might have issues with small values of \( k \), such as \( k = 1 \) or \( k = 2 \), where the cycle of Fibonacci numbers modulo \( k \) is very short.
3. **Edge Cases**: Test cases where \( n \) and \( k \) have specific properties that might cause the code to fail, such as when \( k \) is a prime number or when \( n \) is a small value.

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

```plaintext
3
1 1
2 2
1000000000000 1377
```

### Explanation:
1. **Test Case 1**: \( n = 1 \), \( k = 1 \)
   - The first Fibonacci number that is divisible by 1 is \( f(1) = 1 \). The expected output is 1.
2. **Test Case 2**: \( n = 2 \), \( k = 2 \)
   - The second Fibonacci number that is divisible by 2 is \( f(3) = 2 \). The expected output is 3.
3. **Test Case 3**: \( n = 1000000000000 \), \( k = 1377 \)
   - This is a large value of \( n \) and a moderate value of \( k \). The code should handle this efficiently and correctly.

This test input should help in identifying any issues in the provided code, especially in handling large values of \( n \) and small values of \( k \).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
import java.util.Scanner;

public class FibonacciCount {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        long div = 1000000007; // 1e9 + 7
        int t = scanner.nextInt();

        while (t-- > 0) {
            long n = scanner.nextLong();
            long k = scanner.nextLong();

            if (k == 1) {
                System.out.println(n);
                continue;
            }

            long x = 2;
            long prv1 = 1, prv2 = 1;
            long ct = 3;

            while (x % k != 0) {
                ++ct;
                prv2 = prv1 % k;
                prv1 = x % k;
                x = prv1 + prv2;
                x = x % k;
            }

            long ans = ((n % div) * (ct % div)) % div;

            System.out.println(ans);
        }

        scanner.close();
    }
}

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

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

Think step by step.

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

1. **Handling large values of `n` and `k`**: The code uses a loop to find the index of the first Fibonacci number divisible by `k`, which can be inefficient for large `k`.
2. **Modulo operation**: The code applies the modulo operation incorrectly, especially when calculating the final result.

Let's consider the following edge cases:

1. **Small values of `k`**: The code might handle small values of `k` correctly, but it's still worth testing.
2. **Large values of `k`**: The code might fail to find the correct index within a reasonable time for large `k`.
3. **Large values of `n`**: The code might fail to handle large values of `n` correctly due to the modulo operation.
4. **Edge case where `k` is a large prime number**: This can make the loop run for a very long time.

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

```plaintext
3
1 1
1000000000000 1
1000000000000 100000
```

This test input includes:

1. A simple case where `k` is 1, which should return `n` directly.
2. A large value of `n` with `k` as 1, which should also return `n` directly.
3. A large value of `n` with a large value of `k`, which can expose issues in the loop and modulo operations.

This test input is small and complete, making it suitable for testing the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
def repeat_pattern(k):
    first, second = 0, 1
    for i in range(0, k * k):
        first, second = second, (first + second) % k
        if first == 0:
            if second == 1:
                return i+1

def mod_series(k, n):
    length = repeat_pattern(k)
    fib_mod = [0, 1]

    for i in range(2, length):
        fib_mod.append((fib_mod[-1] + fib_mod[-2]) % k)

    return fib_mod

def count_divisible_indices(k, length, fib_mod):
    indices = []
    for i in range(length):
        if fib_mod[i] == 0:
            indices.append(i + 1)
    return indices

def fibonacci_index_divisible_by_k(n, k):
    length = repeat_pattern(k)
    fib_mod = mod_series(k, length)

    all_ind = count_divisible_indices(k, length, fib_mod)

    count = len(all_ind)

    complete_periods = n // count
    remainder = n % count
    if remainder == 0:
        return all_ind[-1] + (complete_periods - 1) * length
    else:
        return all_ind[remainder - 1] + complete_periods * length

MOD  = 10**9 + 7
t = int(input())
while(t):
    t-=1
    n,k = map(int, input().split())
    if k == 1:
        print(n)
        continue
    result = fibonacci_index_divisible_by_k(n+1, k)
    print((result-1)%MOD)

**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
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
m=10**9+7

for _ in range(int(input())):
    z,x=map(int,input().split())
    q=1
    w=0
    t=1
    if x==1:
        print(z)
    else:
        while True:
            w,q=q,q+w
            t+=1
            if q%x==0:
                k=z*t
                break
        print(k%m)

**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 the following aspects:

1. **Edge Cases**: Test the boundaries of the input constraints, such as the smallest and largest possible values for \( n \) and \( k \).
2. **Special Cases**: Test specific values that might cause issues, such as \( k = 1 \) or \( n = 1 \).
3. **Performance**: Test large values of \( n \) and \( k \) to ensure the code handles large inputs efficiently.

Given the constraints:
- \( 1 \le t \le 10^4 \)
- \( 1 \le n \le 10^{18} \)
- \( 1 \le k \le 10^5 \)

Let's create a small-scale test input that includes a mix of edge cases and performance tests:

```plaintext
4
1 1
2 2
1000000000000 1
1000000000000 100000
```

### Explanation:
1. **Case 1: \( n = 1, k = 1 \)**:
   - This tests the smallest possible values for \( n \) and \( k \).
   - Expected output: 1

2. **Case 2: \( n = 2, k = 2 \)**:
   - This tests the second smallest values for \( n \) and \( k \).
   - Expected output: 3

3. **Case 3: \( n = 10^{18}, k = 1 \)**:
   - This tests a very large value for \( n \) and the smallest possible value for \( k \).
   - Expected output: 1000000000000 (mod \( 10^9 + 7 \)) = 1000000000000

4. **Case 4: \( n = 10^{18}, k = 100000 \)**:
   - This tests a very large value for \( n \) and the largest possible value for \( k \).
   - Expected output: This is a more challenging case to compute, but it should still be handled correctly by the algorithm.

This test input should help expose any issues in the buggy code, particularly with handling large values and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
import java.io.*;
import java.util.*;
public class Main {
    static FastReader fr =  new FastReader();
    static PrintWriter out = new PrintWriter(System.out);
    static long [] fact;

    public static void Factorial(int n){
        fact = new long [n+1];

        fact[0] = 1l;
        fact[1] = 1l;
        for(int i=2; i<=n; i++){
            fact[i] = ModMul(fact[i-1], i);
        }
    }
    public static void Solve() throws IOException {
        long Seq [] = new long [(int)1e5+1];
        int q = fr.nextInt();
        while (q-- > 0){
            long a = fr.nextLong();
            long b = fr.nextLong();

            if(b==1){
                out.println(a);
                continue;
            }
            int idx=0;
            long prev=1l, prev2=1l;
            for(int i=3; i<=(int)1e5; i++){
                long curr = ((prev%b) + (prev2%b)) % b;

                if(curr==0l) {
                    idx = i;
                    break;
                }
                prev = prev2;
                prev2 = curr;
            }
            out.println(ModMul(idx, a%MOD));
        }
    }

    static int CustomCompare(long a, long b){
        if(a <= b) return -1;
        return 1;
    }

    static long BinaryExpo(long x, long y) {
        if (y == 0) return 1;
        long temp = BinaryExpo(x, y / 2);
        temp = ModMul(temp, temp);
        if (y % 2 != 0) return ModMul(temp, x);
        return temp;
    }
    //    -------------------------------(X * Y) % MOD-------------------------------------------------------
    public static long ModMul(long a, long b){
        return ((a % MOD) * (b % MOD)) % MOD;
    }

    //    -------------------------------------------------------------------------------------------------------
    public static void main(String args[]) throws IOException {
//        Factorial((int)1e6);
//        int t = fr.nextInt();
        int t = 1;
        //        long SqxtasrtTime = System.currentTimeMillis();
        while (t-- > 0) {
            Solve();
        }
        //        long endTime = System.currentTimeMillis();
        //        System.out.println("It took " + (endTime - StartTime) + " milliseconds");
        out.close();
    }

    //    --------------------------------FOR TAKING FASTER INPUTS----------------------------------------
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader() { br = new BufferedReader(new InputStreamReader(System.in));}
        String next()
        {
            while (st == null || !st.hasMoreElements()) {
                try { st = new StringTokenizer(br.readLine()); }
                catch (IOException e) { e.printStackTrace(); }
            }
            return st.nextToken();
        }
        int nextInt() { return Integer.parseInt(next()); }
        long nextLong() { return Long.parseLong(next()); }
        double nextDouble() { return Double.parseDouble(next()); }
        String nextLine()
        {
            String str = "";
            try {
                if(st.hasMoreTokens()){str = st.nextToken("\n");}
                else{str = br.readLine();}
            }
            catch (IOException e) {e.printStackTrace();}
            return str;
        }

    }
    static int MOD = (int)(1e9)+7;
    static int maxI = Integer.MAX_VALUE;
    static int minI = Integer.MIN_VALUE;
    static long maxL = Long.MAX_VALUE;
    static long minL = Long.MIN_VALUE;
    public static int max(int a, int b) {return Math.max(a, b);}
    public static int min(int a, int b) {return Math.min(a, b);}
    public static long max(long a, long b) {return Math.max(a, b);}
    public static long min(long a, long b) {return Math.min(a, b);}
}

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

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

Think step by step.

Response:
```plaintext
3
1 1
1000000000000 1
1000000000000 1377
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
//  ______________________________________HAR HAR MAHADEV______________________________________

import java.lang.reflect.Array;
import java.util.*;
import java.io.*;
import java.math.*;

public class Codechef {

    static int mod = (int) 1e9 + 7;

        static void solve(int cct, int tcc) throws IOException {
        long n=in.nextLong();
        long k=in.nextLong();
        if(k==1)
        {
            out.println(n);
            return;
        }
        long f=1;
        long s=1;
        long div;
        int i=3;
        while(true)
        {
            long nf= f+s;
            nf%=k;
            if(nf==0)
            {
                div=i;
                break;
            }
            f=s;
            s=nf;
            i++;
        }
            out.println(((n%mod)*(div%mod))%mod);

}

    // *-----------------------------------MAIN--------------------------------------------//
    public static void main(String[] args) throws Exception {
//      seive();
//        fact();
        int t = 1, ct = 1;
        t = in.nextInt();
        while (t-- != 0) {
            solve(ct, t);
        }
        out.close();
    }

    static class Reader {
        BufferedReader in;
        StringTokenizer st;

        public Reader() {
            in = new BufferedReader(new InputStreamReader(System.in));
            st = new StringTokenizer("");

        }

        public String nextLine() throws IOException {
            st = new StringTokenizer("");
            return in.readLine();
        }

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

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

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

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

    static Reader in = new Reader();
    static PrintWriter out = new PrintWriter(System.out);

}

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

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

Think step by step.

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

1. **Edge Cases for `k`**:
   - When `k = 1`, the code correctly handles this case by directly outputting `n`.
   - For other values of `k`, the code needs to find the first Fibonacci number that is divisible by `k` and then compute the index of the `n`-th such number.

2. **Large Values for `n`**:
   - The code should handle large values of `n` (up to \(10^{18}\)) efficiently.

3. **Special Values for `k`**:
   - Values of `k` that have a small Pisano period (the period of the Fibonacci sequence modulo `k`) can be tricky. For example, if `k = 2`, the Pisano period is 3, and the sequence of Fibonacci numbers modulo 2 is `[0, 1, 1, 0, 1, 1, ...]`.

4. **Modulo Operation**:
   - The code should correctly handle the modulo operation to ensure the result fits within the required range.

Given these considerations, a challenging test input would include a large value for `n` and a value for `k` that has a small Pisano period, which can cause the loop to run for a long time if not optimized correctly.

Here is a small-scale test input that should expose the bug:

```plaintext
1
1000000000000 2
```

This test case is designed to:
- Use a large value for `n` to test the efficiency of the code.
- Use `k = 2` to test the handling of a small Pisano period.

This input should reveal any issues with the code's ability to handle large values and small Pisano periods efficiently.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
#include<bits/stdc++.h>
#include<unordered_map>
using namespace std;
typedef long long ll;
typedef pair<int, int> PLL;
#define lowbit(x) ((x) & -(x))
const int N = 2e5 + 10;
const int MOD = 1e9 + 7;
ll n,k,pre[2];
vector<ll>s;
map<pair<int,int>,ll>mp;
void solve()
{
	mp.clear();
	s.clear();
	cin>>n>>k;
	if(k==1)
	{
		cout<<n%MOD<<"\n";
		return;
	}
	s.push_back(0);
	s.push_back(0);
	pre[0]=pre[1]=1;
	mp[make_pair(1,1)]=1;
	ll ans=2;
	while(n)
	{
		ans++;
		ans%=MOD;
		ll p=(pre[0]+pre[1])%k;
		s.push_back(s.back()+(p%k==0));
		pre[0]=pre[1],pre[1]=p;
		n-=(p%k==0);
		if(mp.find(make_pair(pre[0],pre[1]))!=mp.end())
		{
			ll t=mp[make_pair(pre[0],pre[1])];
			ll num=n/(s.back()-s[t]);
			n%=(s.back()-s[t]);
			ans=(ans+(s.size()-1-t)%MOD*(num%MOD)%MOD)%MOD;
			if(n==0)
			{
				for(int i=s.size()-1;;i--)
				{
					if(s[i]==s[i-1])
					{
						ans--;
						if(ans<0)
							ans+=MOD;
					}
					else
						break;
				}
			}
			else
			{
				for(int i=t+1;n;i++)
				{
					ans++;
					ans%=MOD;
					n-=(s[i]!=s[i-1]);
				}
			}
			break;
		}
		mp[make_pair(pre[0],pre[1])]=s.size()-1;
	}
	cout<<ans<<"\n";
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	int T = 1;
	cin >> T;
	while (T--)
	{
		solve();
		//cout << "\n";
	}
	system("pause");
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that can potentially cause issues. Here are some key points to consider:

1. **Large Values of \( n \) and \( k \)**: The code should handle large values of \( n \) (up to \( 10^{18} \)) and \( k \) (up to \( 10^5 \)) efficiently.
2. **Special Cases for \( k = 1 \)**: The code should correctly handle the case where \( k = 1 \), as this is a common edge case.
3. **Cycles in Fibonacci Sequence Modulo \( k \)**: The code should correctly handle the cycles that occur in the Fibonacci sequence modulo \( k \).
4. **Boundary Conditions**: The code should handle boundary conditions where \( n \) is very small or very large.

Given these considerations, let's construct a small-scale test input that can expose potential bugs:

- **Test Case 1**: \( n = 10^{18} \), \( k = 1 \) (to check the special case handling)
- **Test Case 2**: \( n = 1000000000000 \), \( k = 1377 \) (to check large values)
- **Test Case 3**: \( n = 5 \), \( k = 2 \) (to check small values and cycle detection)
- **Test Case 4**: \( n = 1000000000000000000 \), \( k = 100000 \) (to check the largest possible values)

Here is the test input:

```plaintext
4
1000000000000000000 1
1000000000000 1377
5 2
1000000000000000000 100000
```

This test input includes a mix of large and small values for \( n \) and \( k \), and it should help identify any issues in the code's handling of these cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define N 1000010
#define int long long
#define i128 __int128
#define db long double
#define pii pair<int,int>
#define st first
#define ed second
#define mkp make_pair
#define pb push_back
#define eps 1e-9
#define mod2 998244353
#define mod 1000000007
#define bs 13131
#define bs2 131
#define INF 0x3f3f3f3f3f3f3f3f
#define il inline
#define vi vector<int>
#define ins insert
#define umap unordered_map
#define uset unordered_set
#define R(x) x.begin(),x.end()
#define B(x) x.begin()
#define E(x) x.end()
#define lb lower_bound
#define ub upper_bound
#define prq priority_queue
mt19937_64 rnd(chrono::system_clock::now().time_since_epoch().count());
il int read(){
	int x=0,f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){
		if(ch=='-')f=-1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9'){
		x=(x<<1)+(x<<3)+(ch^48);
		ch=getchar();
	}
	return x*f;
}
int T=1,n,m,q,k;
int f[10000010];
il void solve(){
	map<pii,int> mp;
	n=read();k=read();
	f[1]=1%k;
	int len,pos;
	for(int i=2;1;i++){
		f[i]=(f[i-1]+f[i-2]);if(f[i]>=k)f[i]-=k;
		if(mp[{f[i-1],f[i]}]){
			pos=mp[{f[i-1],f[i]}];
			len=i-mp[{f[i-1],f[i]}];
			break;
		}
		mp[{f[i-1],f[i]}]=i;
	}
	for(int i=1;i<pos;i++){
		if(!f[i])n--;
		if(!n){
			printf("%lld\n",i);
			return ;
		}
	}
	vector<int> p;
	for(int i=pos;i<pos+len;i++)
		if(!f[i])p.pb(i-pos);
	int ans=pos+len*((n-1)/p.size()%mod);
	ans=(ans%mod+p[(n-1)%p.size()])%mod;
	printf("%lld\n",ans);
}
signed main(){
	T=read();
	while(T--)solve();
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider edge cases and scenarios that are likely to cause issues. Here are the steps to think through this:

1. **Edge Cases for \( n \) and \( k \)**:
   - \( n \) can be very large (up to \( 10^{18} \)).
   - \( k \) can be very large (up to \( 10^5 \)).
   - \( n \) can be 1, and \( k \) can be 1 (smallest possible values).
   - \( n \) can be a large number, and \( k \) can be a small number (to test the cycle detection).

2. **Cycle Detection**:
   - The code uses a map to detect cycles in the Fibonacci sequence modulo \( k \). We need to ensure that the cycle detection is robust.
   - A small \( k \) with a large \( n \) can help in testing the cycle detection logic.

3. **Modulo Operations**:
   - The result is taken modulo \( 10^9 + 7 \). We need to ensure that the modulo operations are handled correctly.

4. **Edge Cases for Test Cases**:
   - The number of test cases \( t \) can be 1 (to simplify the input) or a small number to test multiple scenarios.

Given these considerations, let's generate a test input that includes a large \( n \) and a small \( k \) to test the cycle detection and modulo operations.

```plaintext
1
1000000000000 2
```

This test input will:
- Test the cycle detection for a large \( n \) and a small \( k \).
- Ensure that the modulo operations are handled correctly.
- Expose any issues in the cycle detection logic or the handling of large numbers.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Equal XOR Segments
time_limit_ms: 5000
memory_limit_mb: 256
Description: Let us call an array $$$x_1,\dots,x_m$$$ interesting if it is possible to divide the array into $$$k>1$$$ parts so that bitwise XOR of values from each part are equal.

More formally, you must split array $$$x$$$ into $$$k$$$ consecutive segments, each element of $$$x$$$ must belong to exactly $$$1$$$ segment. Let $$$y_1,\dots,y_k$$$ be the XOR of elements from each part respectively. Then $$$y_1=y_2=\dots=y_k$$$ must be fulfilled.

For example, if $$$x = [1, 1, 2, 3, 0]$$$, you can split it as follows: $$$[\color{blue}1], [\color{green}1], [\color{red}2, \color{red}3, \color{red}0]$$$. Indeed $$$\color{blue}1=\color{green}1=\color{red}2 \oplus \color{red}3\oplus \color{red}0$$$.

You are given an array $$$a_1,\dots,a_n$$$. Your task is to answer $$$q$$$ queries:

- For fixed $$$l$$$, $$$r$$$, determine whether the subarray $$$a_l,a_{l+1},\dots,a_r$$$ is interesting.

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 two integers $$$n$$$ and $$$q$$$ ($$$2 \le n \le 2 \cdot 10^5$$$, $$$1 \le q \le 2 \cdot 10^5$$$) — the number of elements in the array and the number of queries respectively.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$0 \le a_i < 2^{30}$$$) — elements of the array.

Each of the next $$$q$$$ lines contains two integers $$$l$$$ and $$$r$$$ ($$$1 \le l < r \le n$$$) describing the query.

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

It is guaranteed that the sum of $$$q$$$ over all testcases does not exceed $$$2 \cdot 10^5$$$.

Output: For each query, output "YES" if the subarray is interesting and "NO" otherwise.

You can output "Yes" and "No" in any case (for example, the strings "yES", "yes", and "Yes" will be recognized as correct answers).

Examples:
input:
4
5 5
1 1 2 3 0
1 5
2 4
3 5
1 3
3 4
5 5
1 2 3 4 5
1 5
2 4
3 5
1 3
2 3
7 4
12 9 10 9 10 11 9
1 5
1 7
2 6
2 7
11 4
0 0 1 0 0 1 0 1 1 0 1
1 2
2 5
6 9
7 11
output:
YES
YES
NO
NO
NO

YES
NO
NO
YES
NO

NO
NO
NO
NO

YES
NO
YES
YES

Note: Explanation for the first test case:

The first query is described in the statement.

In the second query, we should divide $$$[1,2,3]$$$. A possible division is $$$[1,2],[3]$$$, since $$$1\oplus 2=3$$$.

It can be shown that for queries $$$3,4,5$$$, the subarrays are not interesting.

Buggy Code:
#define _CRT_SECURE_NO_WARNINGS 1
#include<iostream>
#include<map>
#include<vector>
#include<queue>
#include<algorithm>
#include<utility>
#include<string>
#include<cstring>
#include<stack>
#include<sstream>
#include<set>
#include<bitset>
#include<cmath>
#include<unordered_map>
using namespace std;
typedef long long ll;
typedef long double ld;
const ll maxn = 2e5 + 10;
const ll maxm = 1e2 + 10;
const ll mod = 1e9 + 7;
const ll mod1 = 1e9 + 7;
const ll mod2 = 1e9 + 9;
ll t;
ll n, m, idx;
ll arr[maxn];
ll sum[maxn];
vector<ll>vec[maxn];
map<ll, ll>mp;
void init()
{
	mp.clear();
	for (ll i = 0; i <= idx; i++)
		vec[i].clear();
	idx = 0;
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin >> t;
	while (t--)
	{
		cin >> n >> m;
		init();
		for (ll i = 1; i <= n; i++)
			cin >> arr[i];
		for (ll i = 1; i <= n; i++)
		{
			sum[i] = sum[i - 1] ^ arr[i];
			if (mp[sum[i]] == 0)
				mp[sum[i]] = ++idx;
			if (arr[i] != 0)
				vec[mp[sum[i]]].push_back(i);
		}
		for (ll i = 1; i <= m; i++)
		{
			ll l, r;
			cin >> l >> r;
			ll now = sum[r] ^ sum[l - 1];
			if (now == 0)
			{
				cout << "YES\n";
				continue;
			}
			now = sum[r];
			if (now == 0)
			{
				ll res1 = 0;
				ll res2 = 0;
				ll idxr = lower_bound(vec[mp[now]].begin(), vec[mp[now]].end(), r) - vec[mp[now]].begin();
				ll idxl = lower_bound(vec[mp[now]].begin(), vec[mp[now]].end(), l) - vec[mp[now]].begin();
				if (arr[r] == 0)
					res1 += idxr - idxl;
				else
					res1 += idxr - idxl + 1;
				if (mp[sum[l-1]] != 0&&sum[l-1]!=0)
				{
					idxr = lower_bound(vec[mp[sum[l - 1]]].begin(), vec[mp[sum[l - 1]]].end(), r) - vec[mp[sum[l - 1]]].begin();
					idxl = lower_bound(vec[mp[sum[l - 1]]].begin(), vec[mp[sum[l - 1]]].end(), l) - vec[mp[sum[l - 1]]].begin();
					res2 += (idxr - idxl);
				}
				if (res1>1&&(res1+res2) % 2&&res2!=0)
					cout << "YES\n";
				else
					cout << "NO\n";
			}
			else
			{
				ll res1 = 0;
				ll res2 = 0;
				ll idxr = lower_bound(vec[mp[now]].begin(), vec[mp[now]].end(), r) - vec[mp[now]].begin();
				ll idxl = lower_bound(vec[mp[now]].begin(), vec[mp[now]].end(), l) - vec[mp[now]].begin();
				if (arr[r] == 0)
					res1 += idxr - idxl;
				else
					res1 += idxr - idxl + 1;
				if (mp[0] != 0)
				{
					idxr = lower_bound(vec[mp[0]].begin(), vec[mp[0]].end(), r) - vec[mp[0]].begin();
					idxl = lower_bound(vec[mp[0]].begin(), vec[mp[0]].end(), l) - vec[mp[0]].begin();
					res2 += (idxr - idxl);
				}
				if (res1>1&&(res1+res2) % 2&&res2!=0)
					cout << "YES\n";
				else
					cout << "NO\n";
			}
		}
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5 5
1 1 2 3 0
1 5
2 4
3 5
1 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Equal XOR Segments
time_limit_ms: 5000
memory_limit_mb: 256
Description: Let us call an array $$$x_1,\dots,x_m$$$ interesting if it is possible to divide the array into $$$k>1$$$ parts so that bitwise XOR of values from each part are equal.

More formally, you must split array $$$x$$$ into $$$k$$$ consecutive segments, each element of $$$x$$$ must belong to exactly $$$1$$$ segment. Let $$$y_1,\dots,y_k$$$ be the XOR of elements from each part respectively. Then $$$y_1=y_2=\dots=y_k$$$ must be fulfilled.

For example, if $$$x = [1, 1, 2, 3, 0]$$$, you can split it as follows: $$$[\color{blue}1], [\color{green}1], [\color{red}2, \color{red}3, \color{red}0]$$$. Indeed $$$\color{blue}1=\color{green}1=\color{red}2 \oplus \color{red}3\oplus \color{red}0$$$.

You are given an array $$$a_1,\dots,a_n$$$. Your task is to answer $$$q$$$ queries:

- For fixed $$$l$$$, $$$r$$$, determine whether the subarray $$$a_l,a_{l+1},\dots,a_r$$$ is interesting.

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 two integers $$$n$$$ and $$$q$$$ ($$$2 \le n \le 2 \cdot 10^5$$$, $$$1 \le q \le 2 \cdot 10^5$$$) — the number of elements in the array and the number of queries respectively.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$0 \le a_i < 2^{30}$$$) — elements of the array.

Each of the next $$$q$$$ lines contains two integers $$$l$$$ and $$$r$$$ ($$$1 \le l < r \le n$$$) describing the query.

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

It is guaranteed that the sum of $$$q$$$ over all testcases does not exceed $$$2 \cdot 10^5$$$.

Output: For each query, output "YES" if the subarray is interesting and "NO" otherwise.

You can output "Yes" and "No" in any case (for example, the strings "yES", "yes", and "Yes" will be recognized as correct answers).

Examples:
input:
4
5 5
1 1 2 3 0
1 5
2 4
3 5
1 3
3 4
5 5
1 2 3 4 5
1 5
2 4
3 5
1 3
2 3
7 4
12 9 10 9 10 11 9
1 5
1 7
2 6
2 7
11 4
0 0 1 0 0 1 0 1 1 0 1
1 2
2 5
6 9
7 11
output:
YES
YES
NO
NO
NO

YES
NO
NO
YES
NO

NO
NO
NO
NO

YES
NO
YES
YES

Note: Explanation for the first test case:

The first query is described in the statement.

In the second query, we should divide $$$[1,2,3]$$$. A possible division is $$$[1,2],[3]$$$, since $$$1\oplus 2=3$$$.

It can be shown that for queries $$$3,4,5$$$, the subarrays are not interesting.

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

using namespace std;
int numTest;
int n, numQueries;
int sum[200005];

int main () {
  std::cin >> numTest;
  for (int testCase = 1; testCase <= numTest; testCase++) {
    std::cin >> n >> numQueries;
    sum[0] = 0;
    for (int i = 1; i <= n; i++) {
      int x;
      std::cin >> x;
      sum[i] = sum[i - 1] ^ x;
    }
    std::map <int, std::vector <int> > position;
    for (int i = 0; i <= n; i++) {
      position[sum[i]].push_back(i);
    }
    for (int query = 1; query <= numQueries; query++) {
      int l, r;
      std::cin >> l >> r;
      if (sum[r] == sum[l - 1]) {
        std::cout << "YES" << std::endl;
        continue;
      }
      std::vector <int> &v = position[sum[r]];
      std::vector <int>::iterator it = std::lower_bound(v.begin(), v.end(), l);
      if (it == v.end()) {
        std::cout << "NO" << std::endl;
        continue;
      }
      int minLeft = *it;
      if (minLeft + 1 >= r) {
        std::cout << "NO" << std::endl;
        continue;
      }
      v = position[sum[l - 1]];
      it = std::lower_bound(v.begin(), v.end(), r);
      if (it == v.begin()) {
        std::cout << "NO" << std::endl;
        continue;
      }
      it--;
      int maxRight = *it;
      if (maxRight <= l) {
        std::cout << "NO" << std::endl;
        continue;
      }
      if (maxRight > minLeft) {
        std::cout << "YES" << std::endl;
      }
      else {
        std::cout << "NO" << std::endl;
      }
    }
    std::cout << std::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
5 5
1 1 2 3 0
1 5
2 4
3 5
1 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Equal XOR Segments
time_limit_ms: 5000
memory_limit_mb: 256
Description: Let us call an array $$$x_1,\dots,x_m$$$ interesting if it is possible to divide the array into $$$k>1$$$ parts so that bitwise XOR of values from each part are equal.

More formally, you must split array $$$x$$$ into $$$k$$$ consecutive segments, each element of $$$x$$$ must belong to exactly $$$1$$$ segment. Let $$$y_1,\dots,y_k$$$ be the XOR of elements from each part respectively. Then $$$y_1=y_2=\dots=y_k$$$ must be fulfilled.

For example, if $$$x = [1, 1, 2, 3, 0]$$$, you can split it as follows: $$$[\color{blue}1], [\color{green}1], [\color{red}2, \color{red}3, \color{red}0]$$$. Indeed $$$\color{blue}1=\color{green}1=\color{red}2 \oplus \color{red}3\oplus \color{red}0$$$.

You are given an array $$$a_1,\dots,a_n$$$. Your task is to answer $$$q$$$ queries:

- For fixed $$$l$$$, $$$r$$$, determine whether the subarray $$$a_l,a_{l+1},\dots,a_r$$$ is interesting.

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 two integers $$$n$$$ and $$$q$$$ ($$$2 \le n \le 2 \cdot 10^5$$$, $$$1 \le q \le 2 \cdot 10^5$$$) — the number of elements in the array and the number of queries respectively.

The next line contains $$$n$$$ integers $$$a_1,\dots,a_n$$$ ($$$0 \le a_i < 2^{30}$$$) — elements of the array.

Each of the next $$$q$$$ lines contains two integers $$$l$$$ and $$$r$$$ ($$$1 \le l < r \le n$$$) describing the query.

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

It is guaranteed that the sum of $$$q$$$ over all testcases does not exceed $$$2 \cdot 10^5$$$.

Output: For each query, output "YES" if the subarray is interesting and "NO" otherwise.

You can output "Yes" and "No" in any case (for example, the strings "yES", "yes", and "Yes" will be recognized as correct answers).

Examples:
input:
4
5 5
1 1 2 3 0
1 5
2 4
3 5
1 3
3 4
5 5
1 2 3 4 5
1 5
2 4
3 5
1 3
2 3
7 4
12 9 10 9 10 11 9
1 5
1 7
2 6
2 7
11 4
0 0 1 0 0 1 0 1 1 0 1
1 2
2 5
6 9
7 11
output:
YES
YES
NO
NO
NO

YES
NO
NO
YES
NO

NO
NO
NO
NO

YES
NO
YES
YES

Note: Explanation for the first test case:

The first query is described in the statement.

In the second query, we should divide $$$[1,2,3]$$$. A possible division is $$$[1,2],[3]$$$, since $$$1\oplus 2=3$$$.

It can be shown that for queries $$$3,4,5$$$, the subarrays are not interesting.

Buggy Code:
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
using namespace std;
typedef long long int ll;
const int N=2e5+10,M=4e5+10;
int n,q;
int a[N];
//struct ii{
//	int u,w;
//}s[N];
int s[N],s_[N];
map<int,vector<int>>m,m_;
int find(int x,int y,int s){
	int q=0,r=m[s].size()-1;
	if(r==-1)return 0x3f3f3f3f;
	while(q<r){
		int mid=(q+r)>>1;
		if(m[s][mid]>=x){
			r=mid;
		}else{
			q=mid+1;
		}
	}
	if(m[s][q]<x||m[s][q]>=y)return 0x3f3f3f3f;
	return m[s][q];
}
int find_(int x,int y,int s){
	int q=0,r=m_[s].size()-1;
	if(r==-1)return -0x3f3f3f3f;
	while(q<r){
		int mid=(q+r)>>1;
		if(m_[s][mid]<=x){
			r=mid;
		}else{
			q=mid+1;
		}
	}
	if(m_[s][q]<=y||m_[s][q]>x)return -0x3f3f3f3f;
	return m_[s][q];
}
void slove(){
	cin>>n>>q;
	m.clear();
	m_.clear();
//	m[0].push_back(0);
	for(int i=1;i<=n;++i){
		cin>>a[i];
		s[i]=s[i-1]^a[i];
		if(a[i])m[s[i]].push_back(i);
//		s[i]={i,s[i-1].w^a[i]};
	}
	s_[n+1]=0;
	for(int i=n;i>=1;i--){
		s_[i]=s_[i+1]^a[i];
		if(a[i])m_[s_[i]].push_back(i);
	}
	while(q--){
		int l,r;
		cin>>l>>r;
		if(s[l-1]==s[r]){
			cout<<"YES"<<endl;
		}else{
				int nl=find(l,r,s[r]);
				int nr=find_(r,l,s_[l-1]);
//				cout<<nl<<' '<<nr<<' ';
			if(nl<nr)cout<<"YES"<<endl;
			else cout<<"NO"<<endl;
		}

	}
}
int main(){
	int T;cin>>T;
	while(T--){
		slove();
	}
}

**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 4
1 2 3 4 5 6
1 3
2 4
3 5
4 6
```