**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Brightness Begins
time_limit_ms: 1000
memory_limit_mb: 256
Description: Imagine you have $n$ light bulbs numbered $1, 2, \ldots, n$. Initially, all bulbs are on. To flip the state of a bulb means to turn it off if it used to be on, and to turn it on otherwise.

Next, you do the following:

  * for each $i = 1, 2, \ldots, n$, flip the state of all bulbs $j$ such that $j$ is divisible by $i^\dagger$.

After performing all operations, there will be several bulbs that are still on. Your goal is to make this number exactly $k$.

Find the smallest suitable $n$ such that after performing the operations there will be exactly $k$ bulbs on. We can show that an answer always exists.

$^\dagger$ An integer $x$ is divisible by $y$ if there exists an integer $z$ such that $x = y\cdot z$.

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

The only line of each test case contains a single integer $k$ ($1 \le k \le 10^{18}$).

Output: For each test case, output $n$ — the minimum number of bulbs.

Examples:
input:
3
1
3
8
output:
2
5
11

Note: In the first test case, the minimum number of bulbs is $2$. Let's denote the state of all bulbs with an array, where $1$ corresponds to a turned on bulb, and $0$ corresponds to a turned off bulb. Initially, the array is $[1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}]$.

In the end, there are $k = 1$ bulbs on. We can also show that the answer cannot be less than $2$.

In the second test case, the minimum number of bulbs is $5$. Initially, the array is $[1, 1, 1, 1, 1]$.

  * After performing the operation with $i = 1$, the array becomes $[\underline{0}, \underline{0}, \underline{0}, \underline{0}, \underline{0}]$.    * After performing the operation with $i = 2$, the array becomes $[0, \underline{1}, 0, \underline{1}, 0]$.    * After performing the operation with $i = 3$, the array becomes $[0, 1, \underline{1}, 1, 0]$.    * After performing the operation with $i = 4$, the array becomes $[0, 1, 1, \underline{0}, 0]$.    * After performing the operation with $i = 5$, the array becomes $[0, 1, 1, 0, \underline{1}]$.

In the end, there are $k = 3$ bulbs on. We can also show that the answer cannot be smaller than $5$.

Buggy Code:
import java.util.Scanner;
public class Brightness{
public static void main(String[] args)  {

    Scanner sc = new Scanner(System.in);
    int test_cases= sc.nextInt();
    for(int i=0 ; i< test_cases ; i++){
        System.out.println(func(sc));
    }
}
static long func(Scanner sc){
    long k= sc.nextLong();
       if(k==1) return 2;
       if(k==2) return 3;

       long x=(long) Math.floor(Math.sqrt(k));
       long ans=k+x;
       long y=(long) Math.floor(Math.sqrt(ans));;
       if(x!=y){
           ans++;
       }

       return 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
4
10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

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

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
// File F.cpp created on 29.09.2024 at 18:29:32
#include <bits/stdc++.h>

using i64 = long long;

#ifdef DEBUG
    #include "../debug.h"
#else
    #define debug(...) void(23)
#endif

template<typename T>
T power(T a, i64 b) {
    T res = 1;
    while(b) {
        if(b & 1) {
            res *= a;
        }
        a *= a;
        b >>= 1;
    }
    return res;
}

constexpr int md = int(1E9) + 7;
struct MInt {
    int val;
    MInt() : val(0) {}
    template<typename T>
    MInt(T v) {
        if(-md <= v && v < md) {
            val = v;
        } else {
            val = v % md;
        }
        if(val < 0) {
            val += md;
        }
    }
    int operator() () { return val; }
    MInt operator+= (MInt rhs) {
        if((val += rhs.val) >= md) {
            val -= md;
        }
        return *this;
    }
    MInt operator-= (MInt rhs) {
        if((val -= rhs.val) < 0) {
            val += md;
        }
        return *this;
    }
    MInt operator*= (MInt rhs) {
        val = (int)(1LL * val * rhs.val % md);
        return *this;
    }
    MInt inv() {
        return power(*this, md - 2);
    }
    MInt operator/= (MInt rhs) {
        return *this *= rhs.inv();
    }
    bool operator== (MInt rhs) {
        return val == rhs.val;
    }
    bool operator!= (MInt rhs) {
        return val != rhs.val;
    }
};
MInt operator+ (MInt lhs, MInt rhs) {
    return lhs += rhs;
}
MInt operator- (MInt lhs, MInt rhs) {
    return lhs -= rhs;
}
MInt operator* (MInt lhs, MInt rhs) {
    return lhs *= rhs;
}
MInt operator/ (MInt lhs, MInt rhs) {
    return lhs /= rhs;
}
std::ostream& operator<< (std::ostream& os, MInt v) {
    return os << v();
}
std::string to_string(MInt x) {
    return to_string(x());
}

using Z = MInt;

struct Comb {
    int n;
    std::vector<Z> _fac, _inv_fac, _inv;
    Comb() : n{0}, _fac{1}, _inv_fac{1}, _inv{0} {}
    Comb(int _n) : Comb() {
        init(_n);
    }

    void init(int m) {
        _fac.resize(m + 1);
        _inv_fac.resize(m + 1);
        _inv.resize(m + 1);

        for(int i = n + 1; i <= m; ++i) {
            _fac[i] = _fac[i - 1] * i;
        }
        _inv_fac[m] = _fac[m].inv();
        for(int i = m; i > n; --i) {
            _inv_fac[i - 1] = _inv_fac[i] * i;
            _inv[i] = _inv_fac[i] * _fac[i - 1];
        }
        n = m;
    }

    Z fac(int m) {
        if(m > n) init(2 * m);
        return _fac[m];
    }
    Z inv_fac(int m) {
        if(m > n) init(2 * m);
        return _inv_fac[m];
    }
    Z inv(int m) {
        if(m > n) init(2 * m);
        return _inv[m];
    }
    Z binom(int m, int r) {
        if(m < r || r < 0) return 0;
        return fac(m) * inv_fac(m - r) * inv_fac(r);
    }
} comb;

std::vector<int> minp, primes;

void sieve(int n) {
    minp.assign(n + 1, 0);
    primes.clear();

    for (int i = 2; i <= n; i++) {
        if (minp[i] == 0) {
            minp[i] = i;
            primes.push_back(i);
        }

        for (auto p : primes) {
            if (i * p > n) {
                break;
            }
            minp[i * p] = p;
            if (p == minp[i]) {
                break;
            }
        }
    }
}

void solve() {
    int N, K, D;
    std::cin >> N >> K >> D;

    const int sqrt_N = int(std::sqrt(N)) + 1;

    sieve(sqrt_N);

    std::vector<int> v;
    for (int l = 1, r; l <= N; l = r + 1) {
        v.emplace_back(N / l);
        r = N / (N / l);
    }

    const int nv = int(v.size());
    auto idx = [&](int x) -> int {
        if (x <= sqrt_N) {
            return nv - x;
        }
        return N / x - 1;
    };

    std::vector<int> f(nv);
    for (int i = 0; i < nv; ++i) {
        f[i] = v[i] - 1;
    }

    const int np = int(std::upper_bound(primes.begin(), primes.end(), sqrt_N) - primes.begin());
    for (int k = 0; k < np; ++k) {
        int p = primes[k];
        for (int i = 0; i < nv && v[i] >= p * p; ++i) {
            f[i] -= f[idx(v[i] / p)] - k;
        }
        debug(p, f);
    }

    debug(v, f);

    std::vector<Z> fp(30);
    for (int i = 1; i < 30; ++i) {
        fp[i] = comb.binom(i * K + D, D);
    }
    debug(fp);
    std::vector<Z> g(nv, 1);
    for (int k = np - 1; k >= 0; --k) {
        int p = primes[k];
        for (int i = 0; i < nv && p <= v[i]; ++i) {
            i64 pw = p;
            for (int e = 1; pw <= v[i]; ++e, pw *= p) {
                g[i] += g[idx(int(v[i] / pw))] * fp[e];
            }
        }
        debug(p, g);
    }

    Z ans = g[0];
    debug(ans);
    for (int i = 0; i < nv; ++i) {
        if (v[i] <= sqrt_N) {
            break;
        }
        debug(g[nv - i - 1], f[i] - f[i + 1]);
        ans += g[nv - i - 1] * (f[i] - f[i + 1]) * fp[1];
    }

    std::cout << ans << '\n';

    return;
}

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

    int TT = 1; std::cin >> TT;
    for (int i = 1; i <= TT; ++i) {
        solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to think through this:

1. **Edge Cases for \( n \):**
   - Small values of \( n \) (e.g., 1, 2, 3).
   - Values of \( n \) that are powers of small primes (e.g., 4, 8, 9).
   - Values of \( n \) that are large but within the limit (e.g., \( 10^9 \)).

2. **Edge Cases for \( k \):**
   - Small values of \( k \) (e.g., 1, 2, 3).
   - Large values of \( k \) (e.g., \( 10^5 \)).

3. **Edge Cases for \( d \):**
   - Small values of \( d \) (e.g., 1, 2, 3).
   - Large values of \( d \) (e.g., \( 10^5 \)).

4. **Combining Edge Cases:**
   - Test cases where \( n \), \( k \), and \( d \) are all small.
   - Test cases where \( n \), \( k \), and \( d \) are all large.
   - Test cases where one or two of the parameters are large and the other(s) are small.

Given the constraints and the need for a small-scale test input, let's create a test case that combines various edge cases to maximize the chance of exposing the bug.

Here is the test input:

```plaintext
4
1 1 1
2 1 2
10 1 2
1000000000 100000 100000
```

Explanation:
- **Case 1:** \( n = 1 \), \( k = 1 \), \( d = 1 \)
  - Simplest case, should be straightforward.
- **Case 2:** \( n = 2 \), \( k = 1 \), \( d = 2 \)
  - Small \( n \) with a small \( d \), but slightly more complex than the first case.
- **Case 3:** \( n = 10 \), \( k = 1 \), \( d = 2 \)
  - Small \( n \) with a small \( d \), but a bit larger \( n \).
- **Case 4:** \( n = 10^9 \), \( k = 10^5 \), \( d = 10^5 \)
  - Large values for all parameters, to test the limits of the algorithm.

This test input should help in identifying any issues with the buggy code, especially in handling large inputs and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

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

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

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

using namespace std;

const int MOD = 1e9 + 7;

int f(int n, int d)
{
    if (d == 0) return 1;
    int res = 0;
    for (int i = 1; i * i <= n; ++i)
    {
        if (n % i == 0) {
            res += f(i, d - 1);
            if (i * i != n) res += f(n / i, d - 1);
            res %= MOD;
        }
    }
    return res;
}

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n, k, d;
        cin >> n >> k >> d;
        int ans = 0;
        for (int i = 1; i <= n; ++i)
        {
            ans = (ans + f(pow(i, k), d)) % MOD;
        }
        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 create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can stress the algorithm. Here are the key considerations:

1. **Large Input Values**: The code might fail with large values of `n`, `k`, and `d` due to excessive recursion or time complexity.
2. **Edge Cases**: Values like `n = 1`, `k = 1`, `d = 1` should be tested as they often expose simple but critical bugs.
3. **Special Numbers**: Numbers with many divisors (e.g., highly composite numbers) can be particularly challenging.
4. **Multiple Test Cases**: Ensure the test input includes multiple test cases to cover different scenarios.

Given these considerations, a small-scale but challenging test input could be:

```plaintext
4
1000000000 1 1
1000000000 1 10
1 100000 100000
10 100000 100000
```

Explanation:
1. **Test Case 1**: `1000000000 1 1`
   - This tests the case where `n` is very large, and `d` is 1. The function should return the sum of divisors of all numbers from 1 to 1000000000, which is a large number.
2. **Test Case 2**: `1000000000 1 10`
   - This tests the case where `n` is very large, and `d` is 10. The function should handle deep recursion and large numbers of divisors.
3. **Test Case 3**: `1 100000 100000`
   - This tests the case where `n` is 1, `k` is very large, and `d` is very large. The function should handle the edge case of a single number.
4. **Test Case 4**: `10 100000 100000`
   - This tests the case where `n` is small, `k` is very large, and `d` is very large. The function should handle large powers and deep recursion.

This test input will help expose any issues with the recursive function and the handling of large numbers and deep recursion.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

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

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
unordered_map<int, vector<int>> divisorCache;
vector<int> getDivisors(int x) {
    if (divisorCache.find(x) != divisorCache.end()) {
        return divisorCache[x];
    }
    vector<int> divisors;
    for (int i = 1; i * i <= x; ++i) {
        if (x % i == 0) {
            divisors.push_back(i);
            if (i != x / i) {
                divisors.push_back(x / i);
            }
        }
    }
    divisorCache[x] = divisors;
    return divisors;
}
int countLeaves(int num, int depth) {
    if (depth == 0) {
        return 1;
    }

    queue<int> q;
    q.push(num);
    for (int i = 0; i < depth; ++i) {
        int size = q.size();
        while (size--) {
            int x = q.front();
            q.pop();
            vector<int> divisors = getDivisors(x);
            for (int d : divisors) {
                q.push(d);
            }
        }
    }
    return q.size();
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int t;
    cin >> t;

    while (t--) {
        int n, k, d;
        cin >> n >> k >> d;
        long long result = 0;

        for (int i = 1; i <= n; ++i) {
            long long num = pow(i, k);
            int leaves = countLeaves(num, d);
            result = (result + leaves) % MOD;
        }

        cout << result << endl;
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1 100000 100000
1000000000 1 100000
1000000000 100000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

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

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
def get_divisors(n):
    divisors = []
    i = 1
    while i * i <= n:
        if n % i == 0:
            divisors.append(i)
            if i * i != n:
                divisors.append(n // i)
        i += 1
    return sorted(divisors)

def calculate_leaves(n, d, memo=None):
    if memo is None:
        memo = {}

    if d == 0:
        return 1

    if (n, d) in memo:
        return memo[(n, d)]

    divisors = get_divisors(n)
    count = 0
    for div in divisors:
        count += calculate_leaves(div, d - 1, memo)

    memo[(n, d)] = count
    return count

def solve(n, k, d):
    MOD = 10**9 + 7
    memo = {}
    result = 0

    for i in range(1, n + 1):
        num = pow(i, k, MOD)
        result = (result + calculate_leaves(num, d, memo)) % MOD

    return result

def main():
    t = int(input())
    for _ in range(t):
        n, k, d = map(int, input().split())
        print(solve(n, k, d))

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
```plaintext
3
1 100000 100000
1000000000 1 100000
1000000000 100000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

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

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
import sys
from collections import defaultdict

MOD = 10**9 + 7
MAX_N = 10**5

def precompute_divisors(max_n):
    divisors = [[] for _ in range(max_n+1)]
    for i in range(1, max_n + 1):
        for j in range(i, max_n + 1, i):
            divisors[j].append(i)
    return divisors

def precompute_f(max_n, max_d, divisors):
    dp = [[0] * (max_d + 1) for _ in range(max_n + 1)]

    for i in range(1, max_n + 1):
        dp[i][0] = 1

    for depth in range(1, max_d + 1):
        for i in range(1, max_n + 1):
            for div in divisors[i]:
                dp[i][depth] = (dp[i][depth] + dp[div][depth - 1]) % MOD

    return dp

def precompute_powers(max_n, max_k):
    powers = [[1] * (max_k + 1) for _ in range(max_n + 1)]
    for i in range(1, max_n + 1):
        for j in range(1, max_k + 1):
            powers[i][j] = pow(i, j, MAX_N)
    return powers

t = int(input())
queries = [tuple(map(int, input().split())) for _ in range(t)]

max_n = max(n for n, k, d in queries)
max_k = max(k for n, k, d in queries)
max_d = max(d for n, k, d in queries)

divisors = precompute_divisors(MAX_N)
dp = precompute_f(MAX_N, max_d, divisors)
powers = precompute_powers(max_n, max_k)

for n, k, d in queries:
    result = 0
    for i in range(1, n + 1):
        i_pow_k = powers[i][k]
        result = (result + dp[i_pow_k][d]) % MOD
    print(result)

**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
100000 100000 100000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

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

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
import asyncio

divisormemo = {}

async def add_divisor_value(key, value):
    if key in divisormemo:
        divisormemo[key].append(value)
    else:
        divisormemo[key] = [value]

async def Find_Divisors(number):
    if number in divisormemo:
        return divisormemo[number]

    divisors = []
    for i in range(1, int(number**0.5) + 1):
        if number % i == 0:
            await add_divisor_value(number, i)
            divisors.append(i)
            if i != number // i:
                await add_divisor_value(number, number // i)
                divisors.append(number // i)
    return divisors

async def Find_Divisors_iteration(number, iteration):
    current_numbers = [number]
    for _ in range(iteration):
        leaf_divisors = []
        tasks = [Find_Divisors(num) for num in current_numbers]
        results = await asyncio.gather(*tasks)
        for divisors in results:
            leaf_divisors.extend(divisors)
        current_numbers = leaf_divisors
    return len(current_numbers)

async def process_case(LineGrid):
    n, k, d = LineGrid
    sumfunc = 0
    max_nodes = min(n, 100000)  # Limit to 100,000 nodes
    for node in range(1, max_nodes + 1):
        count = await Find_Divisors_iteration(node, d)
        sumfunc += (count ** k)
    return sumfunc

async def main():
    IntCaseCount = int(input())
    if 1 <= IntCaseCount <= 10000:
        tasks = []
        for _ in range(IntCaseCount):
            LineGrid = list(map(int, input().split()))
            tasks.append(process_case(LineGrid))

        results = await asyncio.gather(*tasks)

        # Join results into a single string and print once
        print('\n'.join(map(str, results)))

if __name__ == "__main__":
    asyncio.run(main())

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

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

Think step by step.

Response:
```plaintext
3
1 100000 100000
100000 1 100000
100000 100000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

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

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
import java.util.ArrayList;
import java.util.HashSet;
import java.util.Scanner;
import java.lang.Math;

public class CountingLeaves {

    // Method to find divisors using HashSet to avoid duplicates
    public static HashSet<Integer> findDivisors(int n) {
        HashSet<Integer> divSet = new HashSet<>();
        int sqrtN = (int) Math.sqrt(n);
        for (int i = 1; i <= sqrtN; i++) {
            if (n % i == 0) {
                divSet.add(i);
                divSet.add(n / i);
            }
        }
        return divSet;
    }

    // Method to compute the next level nodes
    public static ArrayList<Integer> nextLevelNodes(ArrayList<Integer> nodes) {
        ArrayList<Integer> result = new ArrayList<>();
        for (int i : nodes) {
            result.addAll(findDivisors(i));
        }
        return result;
    }

    // Method to count leaves at level d
    public static int countLeaves(int n, int d) {
        ArrayList<Integer> nodes = new ArrayList<>(findDivisors(n));
        if (d == 1) {
            return nodes.size();
        }
        while (d > 1) {
            nodes = nextLevelNodes(nodes);
            d--;
        }
        return nodes.size();
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();

        while (t-- > 0) {
            int n = in.nextInt();
            int k = in.nextInt();
            int d = in.nextInt();
            int result = 0;

            // Precompute powers
            if (n == 1){
                System.out.println(1);
                continue;
            }
            for (int i = 1; i <= n; i++) {
                int power = (int) Math.pow(i, k);
                result += countLeaves(power, d);
            }

            System.out.println(result);
        }

        in.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1 1 1
2 1 2
10 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

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

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
import java.util.ArrayList;
import java.util.Scanner;
import java.lang.Math;

public class countLeaves{
    public static ArrayList<Integer> findDivisors(int n) {
        ArrayList<Integer> divlist = new ArrayList<>();
        for(int i = 1; i <=n; i++) {
            if(n%i == 0) {
                divlist.add(i);
            }

        }
        return divlist;
    }
    public static ArrayList<Integer> nextLevelNodes(ArrayList<Integer> nodes) {
        ArrayList<Integer> result = new ArrayList<>();
        for (int i : nodes) {
            result.addAll(findDivisors(i));
        }
        return result;
    }
    public static int countLeafs(int n, int d){

        ArrayList<Integer> nodes;
        ArrayList<Integer> nodes1 = new ArrayList<>();
        nodes = findDivisors(n);
        if (d == 1) {
            return nodes.size();
        }
        while (d > 1) {
            nodes1.addAll(nextLevelNodes(nodes));
            nodes.clear();
            nodes.addAll(nodes1);
            nodes1.clear();
            d--;
        }
        return nodes.size();
    }
    public static void main(String[] args){
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        int n;
        int k;
        int d;
        int result = 0;

        do {
            n = in.nextInt();
            k = in.nextInt();
            d = in.nextInt();
            for(int i = 1; i <= n; i++) {

                result += countLeafs((int)Math.pow(i,k), d);

            }
            System.out.println(result);
            result = 0;
            t--;
        }
        while (t > 0);

        in.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
4
1 1 1
10 1 1
10 1 2
10 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

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

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

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

using namespace std;

template<class F, class S>
ostream &operator<<(ostream &s, const pair<F, S> &v) {
    s << "(" << v.first << ", " << v.second << ")";
    return s;
}
template<ranges::range T> requires (!is_convertible_v<T, string_view>)
istream &operator>>(istream &s, T &&v) {
    for (auto &&x : v) s >> x;
    return s;
}
template<ranges::range T> requires (!is_convertible_v<T, string_view>)
ostream &operator<<(ostream &s, T &&v) {
    for (auto &&x : v) s << x << ' ';
    return s;
}

#ifdef LOCAL
template<class... T> void dbg(T... x) {
    char e{};
    ((cerr << e << x, e = ' '), ...);
}
#define debug(x...) dbg(#x, '=', x, '\n')
#else
#define debug(...) ((void)0)
#endif

#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define ff first
#define ss second

template<class T> inline constexpr T inf = numeric_limits<T>::max() / 2;
bool chmin(auto &a, auto b) { return (b < a and (a = b, true)); }
bool chmax(auto &a, auto b) { return (a < b and (a = b, true)); }

using u32 = unsigned int;
using i64 = long long;
using u64 = unsigned long long;
using i128 = __int128;
using u128 = unsigned __int128;

constexpr i64 mod = 1E9 + 7;

template<size_t N>
struct Comb {
    array<i64, N + 1> fac, ifac, inv;
    Comb() : fac{1, 1}, ifac{1, 1}, inv{0, 1} {
        for (int i = 2; i <= N; i++) {
            inv[i] = inv[mod % i] * (mod - mod / i) % mod;
            fac[i] = fac[i - 1] * i % mod;
            ifac[i] = ifac[i - 1] * inv[i] % mod;
        }
    }
    i64 operator()(int n) {
        return n < 0 ? ifac[-n] : fac[n];
    }
    i64 operator()(int n, int m) {
        if (n < m or m < 0) return 0;
        return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
    }
};
Comb<200005 * 30> comb;

vector<int> primes, minp;
vector<bool> isp;
void sieve(int n) {
    minp.assign(n + 1, 0);
    primes.clear();
    isp.assign(n + 1, 0);
    for (int i = 2; i <= n; i++) {
        if (minp[i] == 0) {
            minp[i] = i;
            isp[i] = 1;
            primes.push_back(i);
        }
        for (i64 p : primes) {
            if (p * i > n) {
                break;
            }
            minp[i * p] = p;
            if (p == minp[i]) {
                break;
            }
        }
    }
}

// Prefix Sums of Multiplicative Functions
// O(N^0.75 / logN)
// calc f(1) + ... + f(N)
// where f is multiplicative function
// construct completely multiplicative functions
// g_i s.t. for all prime x, f(x) = sigma c_i*g_i(x)
// def gsum(x) = g(1) + ... + g(x)
// call apply(g_i, gsum_i, c_i) and call work(f)
struct Min25 {
    const i64 N, sqrtN;
    vector<i64> Q;
    vector<i64> Fp, S;
    int id(i64 x) { return x <= sqrtN ? Q.size() - x : N / x - 1; }
    Min25(i64 N) : N(N), sqrtN(sqrtl(N)) {
        sieve(sqrtN);
        for (i64 l = 1, r; l <= N; l = r + 1) {
            Q.push_back(N / l);
            r = N / (N / l);
        }
        Fp.assign(Q.size(), 0);
        S.assign(Q.size(), 0);
    }
    void apply(const auto &f, const auto &fsum, i64 coef) {
        vector<i64> F(Q.size());
        for (int i = 0; i < Q.size(); i++) {
            F[i] = fsum(Q[i]) - 1;
        }
        for (i64 p : primes) {
            auto t = F[id(p - 1)];
            for (int i = 0; i < Q.size(); i++) {
                if (Q[i] < p * p) {
                    break;
                }
                F[i] -= (F[id(Q[i] / p)] - t) * f(p);
                F[i] %= mod;
            }
        }
        for (int i = 0; i < Q.size(); i++) {
            (Fp[i] += F[i] * coef) %= mod;
        }
    }
    i64 work(const auto &f) {
        S = Fp;
        for (i64 p : primes | views::reverse) {
            i64 t = Fp[id(p)];
            for (int i = 0; i < Q.size(); i++) {
                if (Q[i] < p * p) {
                    break;
                }
                for (i64 pw = p, c = 1; pw * p <= Q[i]; pw *= p, c++) {
                    S[i] += (S[id(Q[i] / pw)] - t) * f(c);
                    S[i] %= mod;
                    S[i] += f(c + 1);
                    S[i] %= mod;
                }
            }
        }
        for (int i = 0; i < Q.size(); i++) {
            S[i]++;
        }
        return S[0];
    }
};

void solve() {
    int n, k, d;
    cin >> n >> k >> d;

    Min25 M(n);
    M.apply(
        [&](i64 x) { return 1; },
        [&](i64 x) { return x; } ,
        d + 1);

    i64 ans = M.work([&](i64 c) {
            return comb(c * k + d, d);
        });

    ans = (ans % mod + mod) % mod;

    cout << ans << '\n';
}

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

    int t = 1;
    cin >> t;

    while (t--) {
        solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can highlight potential issues in the algorithm. Here are the steps to think through this:

1. **Edge Cases**:
   - Small values of \( n \), \( k \), and \( d \) to ensure the algorithm handles basic cases correctly.
   - Large values of \( n \), \( k \), and \( d \) to test the algorithm's performance and correctness under stress.
   - Special values like \( n = 1 \) or \( k = 1 \) or \( d = 1 \) to check if the algorithm handles these cases correctly.

2. **Scenarios to Test**:
   - \( n \) being a prime number.
   - \( n \) being a composite number with multiple prime factors.
   - \( k \) being a small power to ensure the algorithm handles powers correctly.
   - \( d \) being a small value to ensure the algorithm handles small depths correctly.

3. **Combining the Above**:
   - Test cases with \( n \) being a small prime number.
   - Test cases with \( n \) being a small composite number.
   - Test cases with \( n \) being a large composite number.
   - Test cases with \( k \) and \( d \) being small values.

Based on the above considerations, here is a small-scale test input that includes a mix of these cases:

```plaintext
4
1 1 1
2 1 1
4 1 1
10 1 1
```

This test input includes:
- \( n = 1 \), \( k = 1 \), \( d = 1 \) (basic case)
- \( n = 2 \), \( k = 1 \), \( d = 1 \) (small prime number)
- \( n = 4 \), \( k = 1 \), \( d = 1 \) (small composite number)
- \( n = 10 \), \( k = 1 \), \( d = 1 \) (larger composite number)

These cases should help identify any issues in the algorithm, especially in how it handles small and large values, primes, and composites.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

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

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
using pi=pair<int,int>;
#define fi first
#define se second
const int N=1e5+5,M=3e6+5,mod=1e9+7,P=19491001;
int n,k,m,p0,d0,d[N],pr[N*80];
ll f[N];
bitset<N*1000>vis;
ll fc[M],inv[M],val[N];
ll qp(ll a,int b){ll ans=1;for(;b;b>>=1,a=a*a%mod)if(b&1)ans=ans*a%mod;return ans;}
ll C(int n,int m){return n<m||m<0?0:fc[n]*inv[m]%mod*inv[n-m]%mod;}
template<class T1,class T2>
struct Hash{
	int sz,h[P],nxt[N];
    T2 t[N];T1 v[N];
	void clear(){for(int i=1;i<=sz;i++)h[t[i]%P]=0;sz=0;}
    bool count(T1 x) {
        int y=x%P;
		for(int i=h[y];i;i=nxt[i])if(v[i]==x)return 1;
		return 0;
    }
	T2&operator[](T1 x) {
		int y=x%P;
		for(int i=h[y];i;i=nxt[i])if(v[i]==x)return t[i];
		return nxt[++sz]=h[y],h[y]=sz,v[sz]=x,t[sz]=0,t[sz];
	}
};
Hash<int,int>id;
void getp(int n) {
    for(int i=2;i<=n;i++) {
        if(!vis[i])pr[++pr[0]]=i;
        for(int j=1;j<=pr[0]&&pr[j]*i<=n;j++) {
            vis[i*pr[j]]=1;
            if(i%pr[j]==0)break;
        }
    }
}
void init(int n) {
    fc[0]=1;for(int i=1;i<=n;i++)fc[i]=fc[i-1]*i%mod;
    inv[n]=qp(fc[n],mod-2);
    for(int i=n-1;~i;i--)inv[i]=inv[i+1]*(i+1)%mod;
}
int dfs(int x,int lim,int s) {
    int as=s;
    for(int j=x;abs(s)>=pr[j]&&j<=lim;j++)as+=dfs(j+1,lim,-s/pr[j]);
    return as;
}
int get(int n) {
    if(n<=1e8)return upper_bound(pr+1,pr+pr[0]+1,n)-pr-1;
    int lim=pr[0];for(int i=1;i<=pr[0];i++)if(pr[i]*pr[i]>n){lim=i-1;break;}
    return dfs(1,lim,n)+lim-1;
}
void solve() {
    cin>>n>>k>>m,d0=0,id.clear();
    for(int l=1,r;l<=n;l=r+1)r=n/(n/l),d[++d0]=n/l;
    reverse(d+1,d+d0+1);
    for(int i=1;i<=d0;i++)id[d[i]]=i,f[i]=0;
    f[d0]=1;int ct=0;
    for(int i=1;i<=29;i++)val[i]=C(i*k+m,m);
    for(int i=1;pr[i]*pr[i]<=n;i++) {
        for(int j=1;j<=d0;j++)if(f[j]) {
            for(int v=d[j]/pr[i],t=1;v;v/=pr[i],t++) {
                int l=id[v];
                f[l]=(f[l]+f[j]*val[t])%mod;
            }
        }ct++;
    }ll ans=0;
    for(int i=1;i<=d0;i++) {
        ans=(ans+f[i])%mod;
        if(!ct||d[i]>pr[ct])ans=(ans+val[1]*f[i]%mod*(get(d[i])-ct))%mod;
    }cout<<(ans+mod)%mod<<'\n';
}
int main() {
    #ifndef ONLINE_JUDGE
        freopen(".in","r",stdin);
        freopen(".out","w",stdout);
    #endif
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int tt;cin>>tt,init(3e6),getp(1e8);
    while(tt--)solve();
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
100 1 10
1000000000 1 1
1000000000 100000 100000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

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

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
import math

def factors(n):
    ans = []
    for i in range(1, n+1):
        if n % i == 0:
            ans.append(i)
    return ans

#print(factors(6))

def f(n, d):
    theList = [n]
    for i in range(d):
        for index in range(len(theList)):
            theList[index] = factors(theList[index])
        temp = []
        for q in theList:
            for w in q:
                temp += [w]
        theList = temp
        #print(theList)
    return len(theList)

t = int(input())
for o in range(t):
    n, k, d = [int(j) for j in input().split()]
    theAns = 0
    for i in range(1, n+1):
        theAns += f(i ** k, d)
    print(theAns)

**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
100 1 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

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

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
MOD = 10**9 + 7

def precompute_divisors(max_n):
    divisors_count = [0] * (max_n + 1)
    for i in range(1, max_n + 1):
        for j in range(i, max_n + 1, i):
            divisors_count[j] += 1
    return divisors_count

def count_leaves(x, d, dp, divisors_count):
    if dp[x][d] != -1:
        return dp[x][d]

    if d == 0:
        dp[x][d] = 1
        return 1

    total_leaves = 0
    for i in range(1, x + 1):
        if x % i == 0:
            total_leaves += count_leaves(i, d - 1, dp, divisors_count)
            total_leaves %= MOD

    dp[x][d] = total_leaves
    return total_leaves

def solve(n, k, d):
    max_x = int(n**(1/k)) + 1
    divisors_count = precompute_divisors(max_x)

    dp = [[-1] * (d + 1) for _ in range(max_x + 1)]

    total_sum = 0
    for i in range(1, n + 1):
        x = pow(i, k, max_x)
        total_sum += count_leaves(x, d, dp, divisors_count)
        total_sum %= MOD

    return total_sum

# Reading input
import sys
input = sys.stdin.read
data = input().split()

index = 0
t = int(data[index])
index += 1
results = []

for _ in range(t):
    n = int(data[index])
    index += 1
    k = int(data[index])
    index += 1
    d = int(data[index])
    index += 1
    results.append(solve(n, k, d))

# Printing results
for result in results:
    print(result)

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

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

Think step by step.

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

1. **Edge Cases**:
   - Small values of `n`, `k`, and `d`.
   - Large values of `n`, `k`, and `d`.
   - Values of `n` that are powers of small integers.
   - Values of `k` that are large, causing `i^k` to be very large.

2. **Specific Scenarios**:
   - `n = 1` with any `k` and `d`.
   - `k = 1` with any `n` and `d`.
   - `d = 1` with any `n` and `k`.
   - `n` being a prime number.
   - `n` being a composite number with many divisors.

Given these considerations, let's create a small-scale test input that includes a mix of these scenarios:

```plaintext
5
1 1 1
2 1 2
4 2 3
9 3 2
10 1 1
```

This test input includes:
1. A test case with `n = 1`, `k = 1`, `d = 1` to check the base case.
2. A test case with `n = 2`, `k = 1`, `d = 2` to check a small composite number.
3. A test case with `n = 4`, `k = 2`, `d = 3` to check a power of a small integer.
4. A test case with `n = 9`, `k = 3`, `d = 2` to check a larger power of a small integer.
5. A test case with `n = 10`, `k = 1`, `d = 1` to check a composite number.

This test input should be sufficient to expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

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

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
import java.util.Scanner;

import javax.swing.plaf.synth.SynthOptionPaneUI;

public class CountLeaves {
	static int mod = (int) 1e9 +7;
 	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		long[][] dp = new long[11][11];
		int dpd=10 , dpn = 10;
		for(int i =0; i< 11; i++) {
			dp[0][i] =1;
		}
		for(int i=0; i< 11; i++) {
			dp[i][0] =1;
		}
		for(int  i=1; i< 11;i++) {
			for(int j=1; j< 11; j++) {
				for(long k =1; k<= Math.sqrt(j) ; k++ ) {
					if(j%k ==0) {
						dp[i][j] += dp[i-1][(int)(j/k)];
						if(k*k !=j )dp[i][j] += dp[i-1][(int)k];
					}
				}
				//System.out.print(dp[i][j] +" ");
			}
			//System.out.println();
		}
		int t= sc.nextInt();
		while(t-->0) {
			int n= sc.nextInt();
			int k= sc.nextInt();
			int d= sc.nextInt();
			if(n< dpn && d< dpd) {
				int ans =0;
				for(long i=1; Math.pow(i, k)<= n ; i++) {
					ans += dp[d][(int)Math.pow(i, k)];
				}
				System.out.println(ans);
			}
			else {
				dp=makeDp(n, Math.max(dpd, d), dp, dpd, dpn);
				long ans =0;
				dpd = d;
				dpn= n;
				for(long i=1; Math.pow(i, k)<= n ; i++) {
					ans += dp[d][(int)Math.pow(i, k)];
					ans %= mod;
				}
				System.out.println(ans);
			}
		}
	}
	static long[][] makeDp(int n, int d, long[][] cdp, int cdpd, int cdpn){
		long[][] dp = new long[d+1][n+1];
		for(int i=1; i<= n; i++) {
			dp[0][i] =1;
		}
		for(int i=1; i<= d; i++) {
			dp[i][0] =1;
		}
		for(int i=1; i<= d; i++) {
			for(int j=1; j<=n; j++) {
				if(i<= cdpd && j<= cdpn) dp[i][j] = cdp[i][j];
				else {
					for(long k =1; k<= Math.sqrt(j) ; k++ ) {
						if(j%k ==0) {
							dp[i][j] += dp[i-1][(int)(j/k)];
							if(k*k != j )dp[i][j] += dp[i-1][(int)k];
							dp[i][j] %= mod;
						}
					}
				}
				//System.out.print(dp[i][j] +" ");
			}
			//System.out.println();
		}
		return dp;
	}

}

**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
1000000000 100000 100000
1 1 1
10 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

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

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
import java.util.Scanner;

import javax.swing.plaf.synth.SynthOptionPaneUI;

public class CountLeaves {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		long[][] dp = new long[11][11];
		int dpd=10 , dpn = 10;
		for(int i =0; i< 11; i++) {
			dp[0][i] =1;
		}
		for(int i=0; i< 11; i++) {
			dp[i][0] =1;
		}
		for(int  i=1; i< 11;i++) {
			for(int j=1; j< 11; j++) {
				for(long k =1; k<= Math.sqrt(j) ; k++ ) {
					if(j%k ==0) {
						dp[i][j] += dp[i-1][(int)(j/k)];
						if(k*k !=j )dp[i][j] += dp[i-1][(int)k];
					}
				}
				//System.out.print(dp[i][j] +" ");
			}
			//System.out.println();
		}
		int t= sc.nextInt();
		while(t-->0) {
			int n= sc.nextInt();
			int k= sc.nextInt();
			int d= sc.nextInt();
			if(n< dpn && d< dpd) {
				int ans =0;
				for(long i=1; Math.pow(i, k)<= n ; i++) {
					ans += dp[d][(int)Math.pow(i, k)];
				}
				System.out.println(ans);
			}
			else {
				dp=makeDp(n, Math.max(dpd, d), dp, dpd, dpn);
				int ans =0;
				dpd = d;
				dpn= n;
				for(long i=1; Math.pow(i, k)<= n ; i++) {
					ans += dp[d][(int)Math.pow(i, k)];
				}
				System.out.println(ans);
			}
		}
	}
	static long[][] makeDp(int n, int d, long[][] cdp, int cdpd, int cdpn){
		long[][] dp = new long[d+1][n+1];
		for(int i=1; i<= n; i++) {
			dp[0][i] =1;
		}
		for(int i=1; i<= d; i++) {
			dp[i][0] =1;
		}
		for(int i=1; i<= d; i++) {
			for(int j=1; j<=n; j++) {
				if(i<= cdpd && j<= cdpn) dp[i][j] = cdp[i][j];
				else {
					for(long k =1; k<= Math.sqrt(j) ; k++ ) {
						if(j%k ==0) {
							dp[i][j] += dp[i-1][(int)(j/k)];
							if(k*k != j )dp[i][j] += dp[i-1][(int)k];
						}
					}
				}
				//System.out.print(dp[i][j] +" ");
			}
			//System.out.println();
		}
		return dp;
	}

}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on scenarios that might cause issues, such as:

1. **Large Input Values**: The code might fail when dealing with large values of \( n \), \( k \), and \( d \).
2. **Edge Cases**: Values of \( n \), \( k \), and \( d \) that are at the boundary of the input constraints.
3. **Special Cases**: Cases where \( n \) is a perfect power, or \( d \) is 1 or 2, which might not be handled correctly.

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

1. **Large \( n \) and \( d \)**: \( n = 100000 \), \( k = 1 \), \( d = 100000 \)
2. **Edge Case \( d = 1 \)**: \( n = 100000 \), \( k = 1 \), \( d = 1 \)
3. **Edge Case \( n = 1 \)**: \( n = 1 \), \( k = 100000 \), \( d = 100000 \)
4. **Special Case \( n \) as a perfect power**: \( n = 100000 \), \( k = 2 \), \( d = 2 \)

Here is the test input:

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

This test input covers a range of scenarios that can help identify potential 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: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

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

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
c=int(input())
for i in range(c):
    q=[]
    l=[int(o) for o in input().split()]
    for s in range(1,l[0]+1):
        j=[s**l[1]]
        for i in range(l[2]):
            r=j[:]
            for p in r:
                for k in range(1,p):
                    if (p%k==0):
                        j.append(k)
        q.append(len(j))
    print(sum(q)%(10**9+7))

**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 consider the following points:

1. **Edge Cases**: The code might fail when dealing with edge cases such as very small values of \( n \), \( k \), and \( d \).
2. **Large Values**: The code might also fail when dealing with large values of \( n \), \( k \), and \( d \), especially when the number of test cases is large.
3. **Intermediate Values**: Sometimes, intermediate values can cause issues due to the way the code handles the generation of divisors and the tree structure.

Given these considerations, let's create a small-scale test input that includes a mix of these scenarios:

1. A test case with small values of \( n \), \( k \), and \( d \).
2. A test case with large values of \( n \), \( k \), and \( d \).
3. A test case with intermediate values of \( n \), \( k \), and \( d \).

Here is the test input:

```plaintext
3
1 1 1
10 1 2
100 2 3
```

This test input includes:
- A small test case (1 1 1) to check the basic functionality.
- An intermediate test case (10 1 2) to check the handling of a moderate number of leaves.
- A larger test case (100 2 3) to check the performance and correctness with a more complex tree structure.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

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

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
import math
factors_cache = {}
def get_factors(num):
    if num in factors_cache:
        return factors_cache[num]

    factors = set()
    for i in range(1,int(math.sqrt(num))+1):
        if num%i == 0:
            factors.add(i)
            factors.add(num//i)

    factors_cache[num] = sorted(factors)
    return factors_cache[num]

fun_cache = {}
def f(n,d):
    if (n,d) in fun_cache:
        return fun_cache[(n,d)]
    if d==1:
        fun_cache[(n,d)] =  len(get_factors(n))
        return fun_cache[(n,d)]

    ret =  sum([f(i,d-1) for i in get_factors(n)])
    fun_cache[(n,d)] = ret
    return ret

m = 1000000007
def myfun(n,k,d):
    return sum([f(i**k,d)%m for i in range(1,n+1)])%m

T = int(input())
for _ in range(T):
    n, k, d = list(map(int,input().split()))
    print(myfun(n,k,d))

**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 1
2 1 2
10 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Count Leaves
time_limit_ms: 4000
memory_limit_mb: 256
Description: Let $n$ and $d$ be positive integers. We build the the divisor tree $T_{n,d}$ as follows:

  * The root of the tree is a node marked with number $n$. This is the $0$-th layer of the tree.    * For each $i$ from $0$ to $d - 1$, for each vertex of the $i$-th layer, do the following. If the current vertex is marked with $x$, create its children and mark them with all possible distinct divisors$^\dagger$ of $x$. These children will be in the $(i+1)$-st layer.    * The vertices on the $d$-th layer are the leaves of the tree.

For example, $T_{6,2}$ (the divisor tree for $n = 6$ and $d = 2$) looks like this:

![](CDN_BASE_URL/782ea6d6fe622ee5ec986b0d8cae274d)

Define $f(n,d)$ as the number of leaves in $T_{n,d}$.

Given integers $n$, $k$, and $d$, please compute $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

$^\dagger$ In this problem, we say that an integer $y$ is a divisor of $x$ if $y \ge 1$ and there exists an integer $z$ such that $x = y \cdot z$.

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

The only line of each test case contains three integers $n$, $k$, and $d$ ($1 \le n \le 10^9$, $1 \le k,d \le 10^5$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^9$.

Output: For each test case, output $\sum\limits_{i=1}^{n} f(i^k,d)$, modulo $10^9+7$.

Examples:
input:
3
6 1 1
1 3 3
10 1 2
output:
14
1
53

Note: In the first test case, $n = 6$, $k = 1$, and $d = 1$. Thus, we need to find the total number of leaves in the divisor trees $T_{1,1}$, $T_{2,1}$, $T_{3,1}$, $T_{4,1}$, $T_{5,1}$, $T_{6,1}$.

  * $T_{1,1}$ has only one leaf, which is marked with $1$.    * $T_{2,1}$ has two leaves, marked with $1$ and $2$.    * $T_{3,1}$ has two leaves, marked with $1$ and $3$.    * $T_{4,1}$ has three leaves, marked with $1$, $2$, and $4$.    * $T_{5,1}$ has two leaves, marked with $1$ and $5$.    * $T_{6,1}$ has four leaves, marked with $1$, $2$, $3$, and $6$.

The total number of leaves is $1 + 2 + 2 + 3 + 2 + 4 = 14$.

In the second test case, $n = 1$, $k = 3$, $d = 3$. Thus, we need to find the number of leaves in $T_{1,3}$, because $1^3 = 1$. This tree has only one leaf, so the answer is $1$.

Buggy Code:
MOD = 10**9 + 7

def count_leaves(n, d):
    # Memoization dictionary
    memo = {}

    def dp(num, depth):
        if (num, depth) in memo:
            return memo[(num, depth)]

        if depth == 0:
            return 1

        divisors = get_divisors(num)
        result = sum(dp(div, depth - 1) for div in divisors) % MOD
        memo[(num, depth)] = result
        return result

    return dp(n, d)

def get_divisors(num):
    divisors = set()
    for i in range(1, int(num**0.5) + 1):
        if num % i == 0:
            divisors.add(i)
            divisors.add(num // i)
    return list(divisors)

def main():
    import sys
    input = sys.stdin.read
    data = input().split()

    index = 0
    t = int(data[index])
    index += 1

    results = []

    for _ in range(t):
        n = int(data[index])
        k = int(data[index + 1])
        d = int(data[index + 2])
        index += 3

        total_leaves = 0
        for i in range(1, n + 1):
            total_leaves = (total_leaves + count_leaves(pow(i, k), d)) % MOD

        results.append(str(total_leaves))

    print("\n".join(results))

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
```plaintext
3
1 100000 100000
2 100000 100000
3 100000 100000
```