#include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, z, a, b;
    cin >> n >> z;
    vector<int> c(n + 1);
    vector<vector<int>> v(n + 1);
    for (int i = 0; i < n - 1; i++) {
      cin >> a >> b;
      c[a]++, c[b]++;
      v[a].push_back(b);
      v[b].push_back(a);
    }
    if (z == 2)
      cout << (n * (n - 1)) / 2 << '\n';
    else {
      long long int ans = 0;
      for (int i = 1; i <= n; i++) {
        if (c[i] >= z) {
          vector<vector<long long int>> m(n + 1);
          for (int j = 0; j < (int)v[i].size(); j++) {
            queue<pair<int, int>> q;
            vector<int> check(n + 1), cd(n + 1);
            q.push({v[i][j], 1});
            check[v[i][j]] = 1;
            check[i] = 1;
            while (!q.empty()) {
              pair<int, int> atual = q.front();
              q.pop();
              cd[atual.second]++;
              for (int k = 0; k < (int)v[atual.first].size(); k++) {
                int prox = v[atual.first][k];
                if (!check[prox]) {
                  check[prox] = 1;
                  q.push({prox, atual.second + 1});
                }
              }
            }
            for (int k = 1; k <= n; k++) {
              if (cd[k]) {
                m[k].push_back(cd[k]);
              }
            }
          }
          for (int j = 1; j <= n; j++) {
            if ((int)m[j].size() >= z) {
              vector<vector<int>> dp(((int)m[j].size() + 1),
                                     vector<int>(z + 1));
              for (int k = 0; k <= (int)m[j].size(); k++) dp[k][z] = 1;
              for (int k = z - 1; k >= 0; k--) {
                for (int l = (int)m[j].size() - 1; l >= 0; l--) {
                  dp[l][k] += (m[j][l] * dp[l + 1][k + 1]) % M;
                  dp[l][k] = (dp[l][k] + dp[l + 1][k]) % M;
                }
              }
              ans = (ans + dp[0][0]) % M;
            }
          }
        }
      }
      cout << ans << '\n';
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, z, a, b;
    cin >> n >> z;
    vector<int> c(n + 1);
    vector<vector<int>> v(n + 1);
    for (int i = 0; i < n - 1; i++) {
      cin >> a >> b;
      c[a]++, c[b]++;
      v[a].push_back(b);
      v[b].push_back(a);
    }
    if (z == 2)
      cout << (n * (n - 1)) / 2 << '\n';
    else {
      long long int ans = 0;
      for (int i = 1; i <= n; i++) {
        if (c[i] >= z) {
          vector<vector<long long int>> m(n + 1);
          for (int j = 0; j < (int)v[i].size(); j++) {
            queue<pair<int, int>> q;
            vector<int> check(n + 1), cd(n + 1);
            q.push({v[i][j], 1});
            check[v[i][j]] = 1;
            check[i] = 1;
            while (!q.empty()) {
              pair<int, int> atual = q.front();
              q.pop();
              cd[atual.second]++;
              for (int k = 0; k < (int)v[atual.first].size(); k++) {
                int prox = v[atual.first][k];
                if (!check[prox]) {
                  check[prox] = 1;
                  q.push({prox, atual.second + 1});
                }
              }
            }
            for (int k = 1; k <= n; k++) {
              if (cd[k]) {
                m[k].push_back(cd[k]);
              }
            }
          }
          for (int j = 1; j <= n; j++) {
            if ((int)m[j].size() >= z) {
              vector<vector<int>> dp(((int)m[j].size() + 1),
                                     vector<int>(z + 1));
              for (int k = 0; k <= (int)m[j].size(); k++) dp[k][z] = 1;
              for (int k = z - 1; k >= 0; k--) {
                for (int l = (int)m[j].size() - 1; l >= 0; l--) {
                  dp[l][k] += (m[j][l] * dp[l + 1][k + 1]) % M;
                  dp[l][k] = (dp[l][k] + dp[l + 1][k]) % M;
                }
              }
              ans = (ans + dp[0][0]) % M;
            }
          }
        }
      }
      cout << ans << '\n';
    }
  }
  return 0;
}

```