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;
  inline long long read() {
    long long s = 0, w = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') w = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
    return s * w;
  }
  const long long p = 1e9 + 7;
  long long qp(long long x, long long y) {
    long long res = 1;
    for (long long t = x; y; y >>= 1, t = t * t % p)
      if (y & 1) res = res * t % p;
    return res;
  }
  vector<long long> e[1003];
  long long fac[1003], ifac[1003];
  long long C(long long n, long long m) {
    return fac[n] * ifac[n - m] % p * ifac[m] % p;
  }
  long long n, K;
  long long cnt[1003][1003];
  void dfs(long long x, long long fa, long long dep, long long id) {
    ++cnt[id][dep];
    for (long long i : e[x])
      if (i != fa) dfs(i, x, dep + 1, id);
    return;
  }
  long long f[1003];
  signed main() {
    fac[0] = ifac[0] = 1;
    for (long long i = 1; i <= 1000; ++i)
      fac[i] = fac[i - 1] * i % p, ifac[i] = qp(fac[i], p - 2);
    for (long long T = read(); T--;) {
      n = read(), K = read();
      for (long long i = 1; i <= n; ++i) e[i].clear();
      for (long long i = 1; i < n; ++i) {
        long long u = read(), v = read();
        e[u].push_back(v);
        e[v].push_back(u);
      }
      if (K == 2) {
        printf("%lld\n", n * (n - 1) / 2);
        continue;
      }
      long long ans = 0;
      for (long long i = 1; i <= n; ++i) {
        long long c = 0;
        for (long long j = 1; j <= n; ++j)
          for (long long k = 1; k <= n; ++k) cnt[j][k] = 0;
        for (long long j : e[i]) dfs(j, i, 1, ++c);
        for (long long j = 1; j <= n; ++j) {
          for (long long k = 1; k <= n; ++k) f[k] = 0;
          f[0] = 1;
          for (long long k = 1; k <= c; ++k) {
            for (long long l = n; l >= 1; --l)
              f[l] = (f[l] + f[l - 1] * cnt[k][j]) % p;
          }
          ans = (ans + f[K]) % p;
        }
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  