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 MAX_N = 105;
  const int INF = 1000000000;
  const long long MOD = 1000000007LL;
  int n, k;
  vector<int> g[MAX_N];
  int byDist[MAX_N][MAX_N];
  long long dp[MAX_N][MAX_N];
  long long fact[MAX_N];
  long long invFact[MAX_N];
  long long binPow(long long a, long long x) {
    if (x == 0) return 1LL;
    if (x % 2 != 0) return (a * binPow(a, x - 1LL)) % MOD;
    long long halfPow = binPow(a, x / 2LL);
    return (halfPow * halfPow) % MOD;
  }
  void dfs(int u, int p, int d, int startI) {
    byDist[startI][d]++;
    for (int v : g[u])
      if (v != p) dfs(v, u, d + 1, startI);
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    fact[0] = 1;
    for (int i = 1; i < MAX_N; i++) fact[i] = (fact[i - 1] * i) % MOD;
    for (int i = 0; i < MAX_N; i++) invFact[i] = binPow(fact[i], MOD - 2LL);
    int tests;
    cin >> tests;
    while (tests--) {
      cin >> n >> k;
      for (int i = 0; i < n; i++) g[i].clear();
      for (int i = 0; i < n - 1; i++) {
        int u, v;
        cin >> u >> v;
        u--;
        v--;
        g[u].push_back(v);
        g[v].push_back(u);
      }
      if (k == 2) {
        long long c = (fact[n] * invFact[2]) % MOD;
        c = (c * invFact[n - 2]) % MOD;
        cout << c << '\n';
        continue;
      }
      long long result = 0;
      for (int i = 0; i < n; i++) {
        if ((int)g[i].size() < k) continue;
        for (int j = 0; j < (int)g[i].size(); j++)
          for (int d = 0; d < n; d++) byDist[j][d] = 0;
        for (int j = 0; j < (int)g[i].size(); j++) dfs(g[i][j], i, 1, j);
        for (int d = 1; d < n; d++) {
          for (int j = 0; j <= (int)g[i].size(); j++)
            for (int jj = 0; jj <= (int)g[i].size(); jj++) dp[j][jj] = 0;
          dp[0][0] = 1LL;
          for (int j = 0; j < (int)g[i].size(); j++)
            for (int jj = 0; jj < (int)g[i].size(); jj++) {
              dp[j + 1][jj] = (dp[j + 1][jj] + dp[j][jj]) % MOD;
              dp[j + 1][jj + 1] =
                  (dp[j + 1][jj + 1] + dp[j][jj] * byDist[j][d]) % MOD;
            }
          result = (result + dp[(int)g[i].size()][k]) % MOD;
        }
      }
      cout << result << '\n';
    }
    return 0;
  }
  