#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    cin >> n >> k;
    vector<vector<int>> G(n + 1);
    for (int i = 1; i < n; i++) {
      int x, y;
      cin >> x >> y;
      G[x].push_back(y);
      G[y].push_back(x);
    }
    vector<vector<int>> C(n + 1, vector<int>(n + 1));
    const int MOD = (int)1e9 + 7;
    C[0][0] = 1;
    for (int i = 1; i <= n; i++) {
      C[i][0] = C[i][i] = 1;
      for (int j = 1; j < i; j++) {
        C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD;
      }
    }
    vector<int> d(n + 1);
    vector<int> f(n + 1);
    function<void(int, int, int)> dfs = [&](int cur, int fa,
                                            int depth) -> void {
      d[cur] = depth;
      f[cur] = fa;
      for (auto nx : G[cur]) {
        if (nx == fa) {
          continue;
        }
        dfs(nx, cur, depth + 1);
      }
    };
    int ans = 0;
    for (int root = 1; root <= n; root++) {
      dfs(root, 0, 0);
      if ((int)G[root].size() < k) {
        continue;
      }
      vector<vector<int>> cnt(G[root].size(), vector<int>(n));
      function<void(int, vector<int>&)> dfs2 = [&](int cur,
                                                   vector<int>& count) -> void {
        count[d[cur]] += 1;
        for (auto nx : G[cur]) {
          if (nx == f[cur]) {
            continue;
          }
          dfs2(nx, count);
        }
      };
      for (int i = 0; i < (int)G[root].size(); i++) {
        dfs2(G[root][i], cnt[i]);
      }
      for (int depth = 1; depth < n; depth++) {
        vector<int> a(G[root].size());
        for (int i = 0; i < (int)G[root].size(); i++) {
          a[i] = cnt[i][depth];
        }
        vector<int> dp(k + 1);
        dp[0] = 1;
        for (int i = 0; i < (int)G[root].size(); i++) {
          for (int j = k; j >= 1; j--) {
            dp[j] = (dp[j] + 1LL * dp[j - 1] * a[i]) % MOD;
          }
        }
        ans = (ans + dp[k]) % MOD;
      }
    }
    if (k == 2) {
      ans = n * (n - 1) / 2;
    }
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    cin >> n >> k;
    vector<vector<int>> G(n + 1);
    for (int i = 1; i < n; i++) {
      int x, y;
      cin >> x >> y;
      G[x].push_back(y);
      G[y].push_back(x);
    }
    vector<vector<int>> C(n + 1, vector<int>(n + 1));
    const int MOD = (int)1e9 + 7;
    C[0][0] = 1;
    for (int i = 1; i <= n; i++) {
      C[i][0] = C[i][i] = 1;
      for (int j = 1; j < i; j++) {
        C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD;
      }
    }
    vector<int> d(n + 1);
    vector<int> f(n + 1);
    function<void(int, int, int)> dfs = [&](int cur, int fa,
                                            int depth) -> void {
      d[cur] = depth;
      f[cur] = fa;
      for (auto nx : G[cur]) {
        if (nx == fa) {
          continue;
        }
        dfs(nx, cur, depth + 1);
      }
    };
    int ans = 0;
    for (int root = 1; root <= n; root++) {
      dfs(root, 0, 0);
      if ((int)G[root].size() < k) {
        continue;
      }
      vector<vector<int>> cnt(G[root].size(), vector<int>(n));
      function<void(int, vector<int>&)> dfs2 = [&](int cur,
                                                   vector<int>& count) -> void {
        count[d[cur]] += 1;
        for (auto nx : G[cur]) {
          if (nx == f[cur]) {
            continue;
          }
          dfs2(nx, count);
        }
      };
      for (int i = 0; i < (int)G[root].size(); i++) {
        dfs2(G[root][i], cnt[i]);
      }
      for (int depth = 1; depth < n; depth++) {
        vector<int> a(G[root].size());
        for (int i = 0; i < (int)G[root].size(); i++) {
          a[i] = cnt[i][depth];
        }
        vector<int> dp(k + 1);
        dp[0] = 1;
        for (int i = 0; i < (int)G[root].size(); i++) {
          for (int j = k; j >= 1; j--) {
            dp[j] = (dp[j] + 1LL * dp[j - 1] * a[i]) % MOD;
          }
        }
        ans = (ans + dp[k]) % MOD;
      }
    }
    if (k == 2) {
      ans = n * (n - 1) / 2;
    }
    cout << ans << "\n";
  }
  return 0;
}

```