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;
  const int N = 2e2 + 5, mod = 1e9 + 7;
  int e[N], ne[N], h[N], idx;
  int d[N];
  long long f[N][N];
  void add(int a, int b) { e[idx] = b, ne[idx] = h[a], h[a] = idx++; }
  void dfs(int x, int fa, int dis, int f) {
    d[dis]++;
    for (int i = h[x]; ~i; i = ne[i]) {
      int u = e[i];
      if (u == fa) continue;
      dfs(u, x, dis + 1, f);
    }
  }
  int main() {
    int _;
    cin >> _;
    memset(h, -1, sizeof h);
    while (_--) {
      int n, k;
      cin >> n >> k;
      for (int i = 1; i < n; i++) {
        int a, b;
        cin >> a >> b;
        add(a, b);
        add(b, a);
      }
      long long ans = 0;
      if (k == 2) {
        cout << n * (n - 1) / 2 << endl;
      } else {
        vector<int> g[N];
        for (int i = 1; i <= n; i++) {
          for (int i = 1; i <= n; i++) g[i].clear();
          int x = i;
          for (int j = h[x]; ~j; j = ne[j]) {
            memset(d, 0, sizeof d);
            int u = e[j];
            dfs(u, x, 1, x);
            for (int i = 1; i <= n; i++) g[i].push_back(d[i]);
          }
          for (int i = 1; i <= n; i++) {
            f[0][0] = 1;
            if (g[i].size() < k) continue;
            for (int j = 1; j <= g[i].size(); j++) {
              for (int w = 0; w <= k; w++) {
                f[j][w] = f[j - 1][w];
                if (w >= 1)
                  f[j][w] = (f[j][w] + f[j - 1][w - 1] * g[i][j - 1] % mod) % mod;
              }
            }
            ans = (ans + f[g[i].size()][k]) % mod;
          }
        }
        cout << ans << endl;
      }
      idx = 0;
      for (int i = 1; i <= n; i++) h[i] = -1;
    }
  }
  