#include <bits/stdc++.h>
using namespace std;
long long t;
long long k, n;
long long u_, v_;
vector<long long> g[107];
long long d[107][107];
long long f[107][107];
long long v[107];
long long dfs(long long root, long long cur, long long dis) {
  d[root][dis]++;
  v[cur] = 1;
  for (auto nxt : g[cur]) {
    if (v[nxt] == 0) {
      dfs(root, nxt, dis + 1);
    }
  }
}
signed main() {
  cin >> t;
  while (t--) {
    cin >> n >> k;
    long long ans = 0;
    for (long long i = 1; i <= n; ++i) {
      g[i].clear();
    }
    for (long long i = 1; i <= n - 1; ++i) {
      cin >> u_ >> v_;
      g[u_].push_back(v_);
      g[v_].push_back(u_);
    }
    if (k == 2) {
      cout << ((n * (n - 1)) / 2) << endl;
      continue;
    }
    for (long long i = 1; i <= n; ++i) {
      if (g[i].size() >= k) {
        memset(d, 0, sizeof(d));
        memset(v, 0, sizeof(v));
        v[i] = 1;
        for (long long j = 0; j < g[i].size(); ++j) {
          dfs(j + 1, g[i][j], 1);
        }
        for (long long dep = 1; dep <= n; dep++) {
          for (long long j = 0; j <= g[i].size(); j++) {
            f[j][0] = 1;
          }
          for (long long j = 1; j <= g[i].size(); j++) {
            for (long long k = 1; k <= j; k++) {
              f[j][k] =
                  (f[j - 1][k] + (f[j - 1][k - 1] * d[j][dep]) % 1000000007) %
                  1000000007;
            }
          }
          ans += f[g[i].size()][k];
          ans %= 1000000007;
        }
      }
    }
    cout << ans << endl;
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
long long t;
long long k, n;
long long u_, v_;
vector<long long> g[107];
long long d[107][107];
long long f[107][107];
long long v[107];
long long dfs(long long root, long long cur, long long dis) {
  d[root][dis]++;
  v[cur] = 1;
  for (auto nxt : g[cur]) {
    if (v[nxt] == 0) {
      dfs(root, nxt, dis + 1);
    }
  }
}
signed main() {
  cin >> t;
  while (t--) {
    cin >> n >> k;
    long long ans = 0;
    for (long long i = 1; i <= n; ++i) {
      g[i].clear();
    }
    for (long long i = 1; i <= n - 1; ++i) {
      cin >> u_ >> v_;
      g[u_].push_back(v_);
      g[v_].push_back(u_);
    }
    if (k == 2) {
      cout << ((n * (n - 1)) / 2) << endl;
      continue;
    }
    for (long long i = 1; i <= n; ++i) {
      if (g[i].size() >= k) {
        memset(d, 0, sizeof(d));
        memset(v, 0, sizeof(v));
        v[i] = 1;
        for (long long j = 0; j < g[i].size(); ++j) {
          dfs(j + 1, g[i][j], 1);
        }
        for (long long dep = 1; dep <= n; dep++) {
          for (long long j = 0; j <= g[i].size(); j++) {
            f[j][0] = 1;
          }
          for (long long j = 1; j <= g[i].size(); j++) {
            for (long long k = 1; k <= j; k++) {
              f[j][k] =
                  (f[j - 1][k] + (f[j - 1][k - 1] * d[j][dep]) % 1000000007) %
                  1000000007;
            }
          }
          ans += f[g[i].size()][k];
          ans %= 1000000007;
        }
      }
    }
    cout << ans << endl;
  }
}

```