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;
vector<int> g[101];
vector<long long> v[101];
bool check[101];
long long c[101][101];
void f(int root, int parent, int k) {
  if (!check[k]) {
    v[k].push_back(1);
    check[k] = true;
  } else
    v[k].back()++;
  for (int i = 0; i < g[root].size(); i++)
    if (g[root][i] != parent) f(g[root][i], root, k + 1);
}
int main() {
  c[0][0] = 1;
  for (int i = 1; i <= 100; i++) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; j++)
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % 1000000007;
  }
  int t, n, k, a, b;
  long long ans, temp;
  scanf("%d", &t);
  for (; t > 0; t--) {
    scanf("%d %d", &n, &k);
    for (int i = 1; i < n; i++) {
      scanf("%d %d", &a, &b);
      g[a].push_back(b);
      g[b].push_back(a);
    }
    if (k == 1)
      printf("%d\n", n);
    else if (k == 2)
      printf("%lld\n", (long long)n * (long long)(n - 1) / 2 % 1000000007);
    else {
      ans = 0;
      for (int i = 1; i <= n; i++) {
        for (int j = 0; j < g[i].size(); j++) {
          memset(check, false, sizeof(check));
          f(g[i][j], i, 1);
        }
        for (int j = 1; j <= n; j++) {
          if ((int)v[j].size() >= k) {
            temp = 1;
            for (int r = 0; r < v[j].size(); r++)
              temp = temp * v[j][r] % 1000000007;
            temp = temp * c[v[j].size()][k];
            ans = (ans + temp) % 1000000007;
          }
          v[j].clear();
        }
      }
      printf("%lld\n", ans);
    }
    for (int i = 1; i <= n; i++) g[i].clear();
  }
  return 0;
}
