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 = 521, mod = 1e9 + 7;
  int n, K, cnt, mxdep, tot;
  int v[N], dep[N][N], f[N][N];
  struct Edge {
    int to, next;
  } e[N << 1];
  void add(int x, int y) {
    e[++tot].to = y;
    e[tot].next = v[x];
    v[x] = tot;
  }
  void dfs(int x, int ff, int depth) {
    ++dep[cnt][depth];
    mxdep = max(mxdep, depth);
    for (int p = v[x]; p; p = e[p].next)
      if (e[p].to != ff) dfs(e[p].to, x, depth + 1);
  }
  int fpow(int x, int y) {
    int res = 1;
    for (; y; y >>= 1, x = 1ll * x * x % mod)
      if (y & 1) res = 1ll * res * x % mod;
    return res;
  }
  int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = x * 10 + ch - '0';
      ch = getchar();
    }
    return x * f;
  }
  int main() {
    int aq = read();
    while (aq--) {
      n = read();
      K = read();
      for (int i = 1; i <= n; i++) v[i] = 0;
      tot = 0;
      for (int x, y, i = 1; i < n; i++)
        x = read(), y = read(), add(x, y), add(y, x);
      if (K == 2) {
        cout << 1ll * n * (n - 1) % mod * fpow(2, mod - 2) % mod << endl;
        continue;
      }
      int ans = 0;
      for (int i = 1; i <= n; i++) {
        cnt = mxdep = 0;
        for (int j = 0; j <= n; j++)
          for (int k = 0; k <= n; k++) dep[j][k] = 0;
        for (int p = v[i]; p; p = e[p].next) ++cnt, dfs(e[p].to, i, 1);
        for (int j = 1; j <= mxdep; j++) {
          f[0][0] = 1;
          for (int k = 1; k <= cnt; k++) {
            f[k][0] = 1;
            for (int t = 1; t <= min(k, K); t++) {
              f[k][t] =
                  (f[k - 1][t] + 1ll * f[k - 1][t - 1] * dep[k][j] % mod) % mod;
            }
          }
          (ans += f[cnt][K]) %= mod;
        }
      }
      cout << ans << endl;
      for (int i = 1; i <= n; i++) v[i] = 0;
      tot = 0;
    }
    return 0;
  }
  