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 maxn = 105;
const int mod = 1e9 + 7;
vector<int> g[maxn];
long long dp[maxn][maxn];
int cnt[maxn];
void dfs(int u, int fa, int d) {
  cnt[d]++;
  for (int i = 0; i < g[u].size(); i++) {
    int j = g[u][i];
    if (j == fa) continue;
    dfs(j, u, d + 1);
  }
}
void solve() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    g[i].clear();
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  long long ans = 0;
  if (k == 2) {
    ans = 1ll * n * (n - 1) / 2;
    ans %= mod;
    cout << ans << endl;
    return;
  }
  for (int i = 1; i <= n; i++) {
    memset(dp, 0, sizeof dp);
    for (int d = 1; d <= n; d++) dp[d][0] = 1;
    for (int j = 0; j < g[i].size(); j++) {
      int v = g[i][j];
      memset(cnt, 0, sizeof(cnt));
      dfs(v, i, 1);
      for (int d = 1; d <= n; d++) {
        for (int q = k; q >= 1; q--) {
          dp[d][q] = (dp[d][q] + cnt[d] * dp[d][q - 1]) % mod;
        }
      }
      for (int d = 1; d <= n; d++) {
        ans += dp[d][k];
        if (ans >= mod) ans -= mod;
      }
    }
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) solve();
}
