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 = 5e2 + 5, INF = 0x3f3f3f3f, mod = 1e9 + 7;
  const double PI = acos(-1), EPS = 1e-8;
  namespace fast_IO {
  long long read() {
    long long num = 0;
    char c;
    bool tag = false;
    while ((c = getchar()) != '-' && c != '+' && (c < '0' || c > '9') && ~c)
      ;
    if (!~c) return EOF;
    if (c == '-')
      tag = true;
    else if (c == '+')
      tag = false;
    else
      num = c ^ 48;
    while ((c = getchar()) >= '0' && c <= '9')
      num = (num << 1) + (num << 3) + (c ^ 48);
    if (tag) return -num;
    return num;
  }
  }  // namespace fast_IO
  vector<int> to[N];
  long long C[N][N], a[N], dp[N][N];
  int q[N];
  bool vis[N];
  void dfs(int x, int step) {
    int head = 0, tail = 0, sum = 1;
    vis[x] = true;
    q[head++] = x;
    while (head != tail) {
      if (--step == 0) {
        if (sum) a[++*a] = sum;
        return;
      }
      int tot = 0;
      while (sum--) {
        int x = q[tail++];
        for (auto k : to[x]) {
          if (vis[k]) continue;
          vis[k] = true;
          ++tot;
          q[head++] = k;
        }
      }
      sum = tot;
    }
  }
  int main() {
    C[0][0] = 1;
    for (int i = 1; i <= 100; ++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 = fast_IO::read();
    while (T--) {
      int n = fast_IO::read(), k = fast_IO::read();
      long long ans = 0;
      for (int i = 1; i <= n; ++i) to[i].clear();
      for (int i = 1; i < n; ++i) {
        int x = fast_IO::read(), y = fast_IO::read();
        to[x].push_back(y);
        to[y].push_back(x);
      }
      if (k > 2) {
        for (int i = 1; i <= n; ++i) {
          if (to[i].size() < k) continue;
          for (int j = 1; j <= n; ++j) {
            *a = 0;
            memset(vis, 0, sizeof(vis));
            vis[i] = true;
            for (auto k : to[i]) dfs(k, j);
            if (*a < k) continue;
            for (int t = 0; t <= *a; ++t) dp[t][0] = 1;
            for (int t = 1; t <= *a; ++t)
              for (int p = 1; p <= t && p <= k; ++p)
                dp[t][p] = (dp[t - 1][p] + dp[t - 1][p - 1] * a[t] % mod) % mod;
            (ans += dp[*a][k]) %= mod;
          }
        }
        printf("%lld\n", ans);
      } else
        printf("%lld\n", C[n][2]);
    }
    return 0;
  }
  