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;
  template <typename _T>
  inline void read(_T &x) {
    x = 0;
    char s = getchar();
    long long f = 1;
    while (s < '0' || s > '9') {
      f = 1;
      if (s == '-') f = -1;
      s = getchar();
    }
    while ('0' <= s && s <= '9') {
      x = (x << 3) + (x << 1) + s - '0';
      s = getchar();
    }
    x *= f;
  }
  const long long np = 105;
  const long long mod = 1e9 + 7;
  long long fac[np], inv[np];
  inline long long power(long long a, long long b) {
    long long res(1);
    while (b) {
      if (b & 1) res = res * a, res %= mod;
      a = a * a;
      a %= mod;
      b >>= 1;
    }
    return res;
  }
  inline long long c(long long n_, long long m_) {
    return ((fac[n_] * inv[n_ - m_] % mod) * inv[m_]) % mod;
  }
  long long head[np], nxt[np * 2], ver[np * 2], tit;
  inline void add(long long x, long long y) {
    ver[++tit] = y;
    nxt[tit] = head[x];
    head[x] = tit;
  }
  long long dep[np], maxn;
  long long sub[np][np];
  long long dp[np][np];
  long long n, k;
  inline void dfs(long long x, long long ff, long long id) {
    dep[x] = dep[ff] + 1;
    maxn = max(dep[x], maxn);
    sub[id][dep[x]]++;
    for (long long i = head[x], v; i; i = nxt[i]) {
      v = ver[i];
      if (v == ff) continue;
      dfs(v, x, id);
    }
  }
  signed main() {
    fac[0] = 1;
    for (long long i = 1; i <= 100; i++) fac[i] = fac[i - 1] * i, fac[i] %= mod;
    for (long long i = 0; i <= 100; i++) inv[i] = power(fac[i], mod - 2);
    long long T;
    read(T);
    while (T--) {
      read(n);
      read(k);
      for (long long i = 1, a, b; i < n; i++) {
        read(a);
        read(b);
        add(a, b);
        add(b, a);
      }
      if (k == 2) {
        long long ans = c(n, 2);
        printf("%lld\n", ans);
        memset(head, 0, sizeof(head));
        tit = 0;
        continue;
      }
      long long Ans = 0;
      for (long long i = 1; i <= n; i++) {
        dep[i] = 1;
        long long ql = 0;
        maxn = 0;
        for (long long q = head[i]; q; q = nxt[q]) dfs(ver[q], i, ++ql);
        if (ql >= k) {
          for (long long h = 1; h <= maxn; h++) {
            dp[0][0] = 1;
            for (long long a = 1; a <= ql; a++)
              for (long long b = 0; b <= k; b++)
                dp[a][b] =
                    (dp[a - 1][b] + (dp[a - 1][b - 1] * sub[a][h]) % mod) % mod;
            Ans += dp[ql][k];
            Ans %= mod;
          }
        }
        memset(sub, 0, sizeof(sub));
        memset(dp, 0, sizeof(dp));
        memset(dep, 0, sizeof(dep));
      }
      printf("%lld\n", Ans);
      for (long long i = 1; i <= n; i++) head[i] = 0;
      tit = 0;
    }
  }
  