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;
using cd = complex<double>;
const double PI = acos(-1);
const int N = 109;
int c[N][N];
int mod = 1e9 + 7;
vector<vector<int>> adj;
void solve() {
  int n, k;
  cin >> n >> k;
  adj = vector<vector<int>>(n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u - 1].push_back(v - 1);
    adj[v - 1].push_back(u - 1);
  }
  if (k == 2) {
    cout << c[n][k] << "\n";
    return;
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    vector<bool> vis(n, 0);
    queue<int> q;
    q.push(i);
    vis[i] = 1;
    for (int sz = 1; !q.empty(); sz = q.size()) {
      ans += c[sz][k];
      if (ans > mod) ans -= mod;
      while (sz--) {
        int cur = q.front();
        q.pop();
        for (int ch : adj[cur]) {
          if (!vis[ch]) {
            vis[ch] = 1;
            q.push(ch);
          }
        }
      }
    }
  }
  cout << ans << "\n";
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  c[0][0] = 1;
  for (int i = 1; i < N; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
  }
  int t;
  cin >> t;
  while (t--) solve();
}
