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 mod = 1000000007;
  inline void add(int &x, int &y) { (x += y) >= mod ? x -= mod : 0; }
  inline int pl(int x, int y) { return (x += y) >= mod ? x -= mod : x; }
  int T, n, m;
  vector<int> E[110];
  int C[110][110];
  int cnt[110][110], tot, dp[110];
  void getdis(int x, int fa, int dis) {
    ++cnt[tot][dis];
    for (auto y : E[x]) {
      if (y == fa) continue;
      getdis(y, x, dis + 1);
    }
  }
  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] = pl(C[i - 1][j - 1], C[i - 1][j]);
    }
    scanf("%d", &T);
    while (T--) {
      scanf("%d %d", &n, &m);
      for (int i = 1, x, y; i < n; ++i) {
        scanf("%d %d", &x, &y);
        E[x].emplace_back(y);
        E[y].emplace_back(x);
      }
      if (m == 2) {
        printf("%d\n", n * (n - 1) / 2);
        for (int i = 1; i <= n; ++i) E[i].clear();
        continue;
      }
      int ans = 0;
      for (int i = 1; i <= n; ++i) {
        for (auto x : E[i]) ++tot, getdis(x, i, 1);
        for (int j = 1; j <= n; ++j) {
          int cnt1 = 0;
          for (int k = 1; k <= tot; ++k) cnt1 += cnt[k][j] != 0;
          if (cnt1 < m) break;
          dp[0] = 1;
          for (int k = 1; k <= tot; ++k)
            for (int l = k; l; --l)
              dp[l] = (dp[l] + 1ll * dp[l - 1] * cnt[k][j]) % mod;
          add(ans, dp[m]);
          memset(dp, 0, tot + 1 << 2);
        }
        for (int j = 1; j <= tot; ++j) memset(cnt[j] + 1, 0, n << 2);
        tot = 0;
      }
      printf("%d\n", ans);
      for (int i = 1; i <= n; ++i) E[i].clear();
    }
    return 0;
  }
  