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 long long md = 1e9 + 7;
int t, n, k, dis[111];
vector<int> v[111];
long long ft[111], iv[111];
long long pw(long long x, long long y) {
  long long rt = 1;
  while (y) {
    if (y & 1) {
      rt *= x;
      rt %= md;
    }
    x *= x;
    x %= md;
    y >>= 1;
  }
  return rt;
}
void dfs(int x, int y, int dd) {
  for (int i = 0; i < v[x].size(); i++) {
    int z = v[x][i];
    if (z == y) {
      continue;
    }
    dis[dd + 1]++;
    dfs(z, x, dd + 1);
  }
}
long long C(long long x, long long y) {
  return ft[x] * iv[y] % md * iv[x - y] % md;
}
int main() {
  ft[0] = 1;
  iv[0] = 1;
  for (int i = 1; i < 105; i++) {
    ft[i] = ft[i - 1] * i % md;
    iv[i] = pw(ft[i], md - 2);
  }
  cin >> t;
  while (t--) {
    cin >> n >> k;
    for (int i = 0; i < n + 5; i++) {
      v[i].clear();
    }
    for (int i = 0; i < n - 1; i++) {
      int a, b;
      cin >> a >> b;
      v[a].push_back(b);
      v[b].push_back(a);
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < n + 5; j++) {
        dis[j] = 0;
      }
      dfs(i, -1, 0);
      for (int j = 1; j <= n; j++) {
        if (dis[j] >= k) {
          ans += C(dis[j], k);
          ans %= md;
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}
