#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int N = 105;
long long f[N][N];
long long cnt[N][N];
vector<int> q[N];
int maxn = 0;
void dfs(int rt, int u, int fa, int dep) {
  cnt[rt][dep]++;
  maxn = max(maxn, dep);
  for (auto &v : q[u]) {
    if (v == fa) continue;
    dfs(rt, v, u, dep + 1);
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    for (int i = 0; i <= n; i++) {
      q[i].clear();
    }
    for (int i = 1; i <= n - 1; i++) {
      int u, v;
      cin >> u >> v;
      q[u].push_back(v);
      q[v].push_back(u);
    }
    if (k == 2) {
      cout << (long long)n * (n - 1) % mod / 2 % mod << '\n';
      continue;
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      memset(cnt, 0, sizeof cnt);
      int mx_d = 0;
      for (auto &u : q[i]) {
        maxn = 0;
        dfs(u, u, i, 0);
        mx_d = max(mx_d, maxn);
      }
      for (int w = 0; w <= mx_d; w++) {
        memset(f, 0, sizeof f);
        f[0][0] = 1;
        f[0][1] = cnt[q[i][0]][w];
        for (int ii = 1; ii < q[i].size(); ii++) {
          f[ii][0] = 1;
          for (int j = 1; j <= min(k, ii + 1); j++)
            f[ii][j] = (f[ii - 1][j] +
                        (long long)f[ii - 1][j - 1] * cnt[q[i][ii]][w] % mod) %
                       mod;
        }
        ans = (ans + f[q[i].size() - 1][k]) % mod;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

A tree is an undirected connected graph without cycles.

You are given a tree of n vertices. Find the number of ways to choose exactly k vertices in this tree (i. e. a k-element subset of vertices) so that all pairwise distances between the selected vertices are equal (in other words, there exists an integer c such that for all u, v (u ≠ v, u, v are in selected vertices) d_{u,v}=c, where d_{u,v} is the distance from u to v).

Since the answer may be very large, you need to output it modulo 10^9 + 7.

Input

The first line contains one integer t (1 ≤ t ≤ 10) — the number of test cases. Then t test cases follow.

Each test case is preceded by an empty line.

Each test case consists of several lines. The first line of the test case contains two integers n and k (2 ≤ k ≤ n ≤ 100) — the number of vertices in the tree and the number of vertices to be selected, respectively. Then n - 1 lines follow, each of them contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v) which describe a pair of vertices connected by an edge. It is guaranteed that the given graph is a tree and has no loops or multiple edges.

Output

For each test case output in a separate line a single integer — the number of ways to select exactly k vertices so that for all pairs of selected vertices the distances between the vertices in the pairs are equal, modulo 10^9 + 7 (in other words, print the remainder when divided by 1000000007).

Example

Input


3

4 2
1 2
2 3
2 4

3 3
1 2
2 3

5 3
1 2
2 3
2 4
4 5


Output


6
0
1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int N = 105;
long long f[N][N];
long long cnt[N][N];
vector<int> q[N];
int maxn = 0;
void dfs(int rt, int u, int fa, int dep) {
  cnt[rt][dep]++;
  maxn = max(maxn, dep);
  for (auto &v : q[u]) {
    if (v == fa) continue;
    dfs(rt, v, u, dep + 1);
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    for (int i = 0; i <= n; i++) {
      q[i].clear();
    }
    for (int i = 1; i <= n - 1; i++) {
      int u, v;
      cin >> u >> v;
      q[u].push_back(v);
      q[v].push_back(u);
    }
    if (k == 2) {
      cout << (long long)n * (n - 1) % mod / 2 % mod << '\n';
      continue;
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      memset(cnt, 0, sizeof cnt);
      int mx_d = 0;
      for (auto &u : q[i]) {
        maxn = 0;
        dfs(u, u, i, 0);
        mx_d = max(mx_d, maxn);
      }
      for (int w = 0; w <= mx_d; w++) {
        memset(f, 0, sizeof f);
        f[0][0] = 1;
        f[0][1] = cnt[q[i][0]][w];
        for (int ii = 1; ii < q[i].size(); ii++) {
          f[ii][0] = 1;
          for (int j = 1; j <= min(k, ii + 1); j++)
            f[ii][j] = (f[ii - 1][j] +
                        (long long)f[ii - 1][j - 1] * cnt[q[i][ii]][w] % mod) %
                       mod;
        }
        ans = (ans + f[q[i].size() - 1][k]) % mod;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

```