#include <bits/stdc++.h>
using namespace std;
const int N = 1e2 + 5;
vector<int> g[N], a[N];
long long ans, cur[N], f[N][N], mod = 1e9 + 7;
void dfs(int u, int p, int l) {
  cur[l]++;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == p) continue;
    dfs(v, u, l + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    ans = 0;
    for (int i = 1; i <= n; i++) {
      g[i].clear();
      cur[i] = 0;
      a[i].clear();
    }
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << endl;
      continue;
    }
    for (int r = 1; r <= n; r++) {
      for (int i = 1; i <= n; i++) {
        a[i].clear();
      }
      if (g[r].size() < k) continue;
      for (int i = 0; i < g[r].size(); i++) {
        for (int i = 1; i <= n; i++) {
          cur[i] = 0;
        }
        dfs(g[r][i], r, 1);
        for (int i = 1; i <= n; i++) {
          a[i].push_back(cur[i]);
        }
      }
      for (int i = 1; i <= n; i++) {
        f[0][0] = 1;
        for (int j = 1; j <= a[i].size(); j++) {
          for (int l = 0; l <= min(k, j); l++) {
            f[j][l] = f[j - 1][l] % mod;
            if (l > 0 && a[i][j - 1] > 0)
              f[j][l] = (f[j][l] + f[j - 1][l - 1] * a[i][j - 1]) % mod;
          }
        }
        ans = (ans + f[a[i].size()][k]) % mod;
      }
    }
    cout << ans << endl;
  }
}

### Prompt

Your challenge is to write a cpp solution to 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 N = 1e2 + 5;
vector<int> g[N], a[N];
long long ans, cur[N], f[N][N], mod = 1e9 + 7;
void dfs(int u, int p, int l) {
  cur[l]++;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == p) continue;
    dfs(v, u, l + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    ans = 0;
    for (int i = 1; i <= n; i++) {
      g[i].clear();
      cur[i] = 0;
      a[i].clear();
    }
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << endl;
      continue;
    }
    for (int r = 1; r <= n; r++) {
      for (int i = 1; i <= n; i++) {
        a[i].clear();
      }
      if (g[r].size() < k) continue;
      for (int i = 0; i < g[r].size(); i++) {
        for (int i = 1; i <= n; i++) {
          cur[i] = 0;
        }
        dfs(g[r][i], r, 1);
        for (int i = 1; i <= n; i++) {
          a[i].push_back(cur[i]);
        }
      }
      for (int i = 1; i <= n; i++) {
        f[0][0] = 1;
        for (int j = 1; j <= a[i].size(); j++) {
          for (int l = 0; l <= min(k, j); l++) {
            f[j][l] = f[j - 1][l] % mod;
            if (l > 0 && a[i][j - 1] > 0)
              f[j][l] = (f[j][l] + f[j - 1][l - 1] * a[i][j - 1]) % mod;
          }
        }
        ans = (ans + f[a[i].size()][k]) % mod;
      }
    }
    cout << ans << endl;
  }
}

```