description: 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
solution: #include <bits/stdc++.h>
  using namespace std;
  int mul(int x, int y) { return (long long)x * y % 1000000007; }
  int add(int x, int y) {
    x += y;
    while (x >= 1000000007) x -= 1000000007;
    while (x < 0) x += 1000000007;
    return x;
  }
  void dfs(int node, int dad, int d, int i, vector<vector<int>>& freq,
           vector<vector<int>>& v) {
    if (freq[i].size() == d) {
      freq[i].push_back(0);
    }
    freq[i][d]++;
    for (int go : v[node]) {
      if (go != dad) {
        dfs(go, node, d + 1, i, freq, v);
      }
    }
  }
  int calc(int node, vector<vector<int>>& v, int k) {
    int n = v.size() >> 1;
    vector<vector<int>> freq(n);
    for (int i = 0; i < v[node].size(); i++) {
      freq[i].push_back(0);
      dfs(v[node][i], node, 1, i, freq, v);
    }
    int res = 0;
    for (int d = 1 + (node < n); d < n * 2; d += 2) {
      vector<vector<int>> dp(v[node].size() + 1, vector<int>(k + 1, 0));
      dp[0][0] = 1;
      for (int id = 1; id <= v[node].size(); id++) {
        int i = id - 1;
        for (int c = 0; c <= k; c++) {
          int val = 0;
          if (freq[i].size() > d) {
            val = freq[i][d];
          }
          dp[id][c] = add(dp[id - 1][c], mul(c > 0 ? dp[id - 1][c - 1] : 0, val));
        }
      }
      res = add(res, dp[v[node].size()][k]);
    }
    return res;
  }
  void solve() {
    int n, k;
    cin >> n >> k;
    vector<vector<int>> v(n << 1);
    for (int i = 0; i < n - 1; i++) {
      int x, y;
      cin >> x >> y;
      --x;
      --y;
      v[x].push_back(n + i);
      v[n + i].push_back(x);
      v[n + i].push_back(y);
      v[y].push_back(n + i);
    }
    if (k == 1) {
      cout << n << "\n";
      return;
    }
    int ans = 0;
    for (int i = 0; i < 2 * n; i++) {
      ans = add(ans, calc(i, v, k));
    }
    cout << ans << "\n";
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  