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>
const long long MAXN = 105;
const long long MOD = 1000000007;
std::vector<long long> Adj[MAXN];
long long d[MAXN][MAXN];
void dfs(long long u, long long p, long long h, long long dep) {
  d[h][dep]++;
  for (long long j : Adj[u]) {
    if (j == p) continue;
    dfs(j, u, h, dep + 1);
  }
}
signed main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  long long t;
  std::cin >> t;
  long long binom[MAXN][MAXN];
  for (long long i = 0; i < MAXN; i++) binom[i][0] = binom[i][i] = 1;
  for (long long j = 2; j < MAXN; j++)
    for (long long k = 1; k < j; k++)
      binom[j][k] = (binom[j - 1][k - 1] + binom[j - 1][k]) % MOD;
  while (t--) {
    long long n, k;
    std::cin >> n >> k;
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < n; j++) d[i][j] = 0;
      Adj[i].clear();
    }
    for (long long i = 1; i < n; i++) {
      long long u, v;
      std::cin >> u >> v;
      u--;
      v--;
      Adj[u].push_back(v);
      Adj[v].push_back(u);
    }
    if (k == 2) {
      std::cout << ((n * (n - 1)) / 2) % MOD << std::endl;
      continue;
    }
    long long ans = 0;
    for (long long i = 0; i < n; i++) {
      dfs(i, -1, i, 0);
      for (long long j = 1; j < n; j++) {
        if (d[i][j] >= k) ans += binom[d[i][j]][k];
        ans %= MOD;
      }
    }
    std::cout << ans << std::endl;
  }
  return 0;
}
