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
#include <bits/stdc++.h>
using namespace std;
const int p = 1000000007;
int t, n, m, x, y, md;
vector<int> a[120];
int c[120];
int f[120][120][120];
void dfs(int x, int y, int d) {
  c[d]++;
  md = max(md, d);
  for (int i : a[x]) {
    if (i != y) {
      dfs(i, x, d + 1);
    }
  }
}
int main() {
  cin >> t;
  for (int tt = 0; tt < t; tt++) {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
      a[i].clear();
    }
    for (int i = 1; i < n; i++) {
      cin >> x >> y;
      a[x].push_back(y);
      a[y].push_back(x);
    }
    if (m == 2) {
      cout << n * (n - 1) / 2 << endl;
      continue;
    }
    memset(f, 0, sizeof f);
    int z = 0;
    for (int i = 1; i <= n; i++) {
      for (int k = 1; k <= n; k++) {
        f[k][0][0] = 1;
      }
      for (int j = 0; j < a[i].size(); j++) {
        memset(c, 0, sizeof c);
        md = 0;
        dfs(a[i][j], i, 1);
        for (int k = 1; k <= md; k++) {
          for (int l = 0; l <= m; l++) {
            f[k][j + 1][l] = f[k][j][l];
            if (l > 0) {
              f[k][j + 1][l] =
                  (f[k][j + 1][l] + (long long)f[k][j][l - 1] * c[k]) % p;
            }
          }
        }
      }
      for (int k = 1; k <= n; k++) {
        z = (z + f[k][a[i].size()][m]) % p;
      }
    }
    cout << z << endl;
  }
  return 0;
}
