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 long long MOD = 1e9 + 7;
  const int N = 128;
  int t, n, k;
  int cnt[N];
  bool vis[N];
  long long dp[N][N];
  vector<int> path[N];
  long long calc(int, int);
  int main() {
    int a, b;
    scanf("%d", &t);
    while (t--) {
      scanf("%d %d", &n, &k);
      for (int i = 0; i <= n; i++) path[i].clear();
      for (int i = 1; i < n; i++) {
        scanf("%d %d", &a, &b);
        path[a].push_back(b);
        path[b].push_back(a);
      }
      if (k == 2) {
        printf("%lld\n", (n - 1ll) * n / 2 % MOD);
        continue;
      }
      long long ans = 0;
      for (int root = 1; root <= n; root++) {
        memset(vis, false, sizeof(vis));
        vis[root] = true;
        int size = path[root].size();
        vector<pair<int, int>> layer;
        for (int i = 0; i < size; i++) {
          int son = path[root][i];
          layer.emplace_back(son, i);
          cnt[i] = 1;
          vis[son] = true;
        }
        while (!layer.empty()) {
          ans = (ans + calc(size, k)) % MOD;
          vector<pair<int, int>> newLayer;
          for (auto node : layer) {
            cnt[node.second]--;
            for (auto son : path[node.first]) {
              if (vis[son]) continue;
              vis[son] = true;
              cnt[node.second]++;
              newLayer.emplace_back(son, node.second);
            }
          }
          layer = newLayer;
        }
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  long long calc(int x, int y) {
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    for (int i = 0; i < x; i++)
      for (int j = 0; j <= y; j++) {
        dp[i + 1][j] = (dp[i + 1][j] + dp[i][j]) % MOD;
        dp[i + 1][j + 1] = (dp[i + 1][j + 1] + dp[i][j] * cnt[i] % MOD) % MOD;
      }
    return dp[x][y];
  }
  