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 maxn = 109;
  const long long mod = 1e9 + 7;
  vector<long long> vec[maxn];
  long long deep[maxn], zi[maxn][maxn], n, k;
  void dfs(long long u, long long father, long long rt) {
    deep[u] = deep[father] + 1;
    zi[rt][deep[u]]++;
    for (auto v : vec[u]) {
      if (v == father) continue;
      dfs(v, u, rt);
    }
  }
  long long f[maxn][maxn], w[maxn];
  void upd(long long &x, long long y) { x = (x + y) % mod; }
  long long DP() {
    for (long long i = 1; i <= w[0]; i++)
      for (long long j = 0; j <= i && j <= k; j++) f[i][j] = 0;
    f[0][0] = 1;
    for (long long i = 1; i <= w[0]; i++)
      for (long long j = 0; j <= i && j <= k; j++) {
        if (j) upd(f[i][j], f[i - 1][j - 1] * w[i] % mod);
        upd(f[i][j], f[i - 1][j]);
      }
    return f[w[0]][k];
  }
  signed main() {
    long long t;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      for (long long i = 1; i < n; i++) {
        long long l, r;
        scanf("%lld%lld", &l, &r);
        vec[l].push_back(r);
        vec[r].push_back(l);
      }
      if (k == 2)
        cout << n * (n - 1) / 2 << endl;
      else {
        long long res = 0;
        for (long long i = 1; i <= n; i++) {
          deep[i] = 0;
          for (auto v : vec[i]) dfs(v, i, v);
          for (long long j = 1; j <= n; j++) {
            w[0] = 0;
            for (auto v : vec[i]) w[++w[0]] = zi[v][j];
            res = (res + DP()) % mod;
          }
          for (auto v : vec[i])
            for (long long j = 1; j <= n; j++) zi[v][j] = 0;
        }
        cout << res << endl;
      }
      for (long long i = 1; i <= n; i++) vec[i].clear();
    }
  }
  