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 N = 105, mod = 1e9 + 7;
  int T, n, m, x, y, soncnt, cnt[N][N], dep[N], son[N];
  long long ans;
  vector<int> G[N];
  long long f[N][N];
  inline void dfs(int x, int fa, int col) {
    dep[x] = dep[fa] + 1, ++cnt[col][dep[x]];
    for (auto y : G[x])
      if (y != fa) dfs(y, x, col);
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> T;
    while (T--) {
      cin >> n >> m, ans = 0;
      for (int i = 1; i <= n; ++i) G[i].clear();
      for (int i = 1; i < n; ++i) {
        cin >> x >> y;
        G[x].push_back(y);
        G[y].push_back(x);
      }
      if (m == 2) {
        cout << (n * (n - 1) / 2) << endl;
        continue;
      }
      for (int i = 1; i <= n; ++i) {
        memset(cnt, 0, sizeof cnt);
        dep[i] = soncnt = 0;
        for (auto y : G[i]) dfs(y, i, y), son[++soncnt] = y;
        for (int D = 1; D <= n; ++D) {
          memset(f, 0, sizeof f);
          f[0][0] = 1;
          for (int j = 1; j <= soncnt; ++j) {
            f[j][0] = f[j - 1][0];
            for (int k = 1; k <= j; ++k)
              f[j][k] =
                  (f[j - 1][k - 1] * cnt[son[j]][D] % mod + f[j - 1][k]) % mod;
          }
          (ans += f[soncnt][m]) %= mod;
        }
      }
      cout << ans << endl;
    }
    return 0;
  }
  