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;
long long n, k, dis[102][102], ans, mod = 1e9 + 7, fact[202];
vector<int> gr[102];
void dfs(int x, int p, int pr, int l) {
  dis[pr][l]++;
  for (int i = 0; i < gr[x].size(); i++) {
    int u = gr[x][i];
    if (u == p) continue;
    dfs(u, x, pr, l + 1);
  }
}
long long pw(long long x, long long y) {
  if (y == 0) return 1;
  long long ans = pw(x, y / 2);
  ans = (ans * ans) % mod;
  if (y % 2) ans = (ans * x) % mod;
  return ans;
}
long long c(long long x, long long y) {
  if (x > y) return 0;
  long long ans = (fact[x] * pw((fact[y] * fact[x - y]) % mod, mod - 2)) % mod;
  return ans;
}
int main() {
  int t;
  cin >> t;
  fact[0] = 1;
  for (int i = 1; i <= 200; i++) fact[i] = (fact[i - 1] * i) % mod;
  while (t--) {
    ans = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j <= 100; j++) dis[i][j] = 0;
      gr[i].clear();
    }
    cin >> n >> k;
    for (int i = 0; i < n - 1; i++) {
      int x, y;
      cin >> x >> y;
      gr[x].push_back(y);
      gr[y].push_back(x);
    }
    for (int i = 1; i <= n; i++) dfs(i, 0, i, 0);
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) ans = (ans + c(dis[i][j], k)) % mod;
    }
    if (k == 2) ans = (ans + n - 1) % mod;
    cout << ans << endl;
  }
}
