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;
  int n, k, c[105][105], d[105], u[105], f[105][105];
  vector<int> e[105];
  void dfs(int x, int fa) {
    d[x] = d[fa] + 1;
    u[d[x]]++;
    for (int y : e[x])
      if (y != fa) dfs(y, x);
  }
  void solve() {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) e[i].clear();
    for (int i = 1; i <= n - 1; i++) {
      int x, y;
      scanf("%d%d", &x, &y);
      e[x].push_back(y);
      e[y].push_back(x);
    }
    if (k == 2) {
      cout << 1LL * n * (n - 1) / 2 << endl;
      return;
    }
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      memset(f, 0, sizeof(f));
      for (int j = 1; j <= n; j++) f[j][0] = 1;
      d[i] = 0;
      for (int y : e[i]) {
        memset(u, 0, sizeof(u));
        dfs(y, i);
        for (int j = 1; j <= n; j++)
          for (int w = k; w >= 1; w--)
            f[j][w] = (f[j][w] + 1LL * u[j] * f[j][w - 1]) % 1000000007;
      }
      for (int j = 1; j <= n; j++) ans = (ans + f[j][k]) % 1000000007;
    }
    printf("%d\n", ans);
  }
  int main() {
    int T;
    scanf("%d", &T);
    while (T--) solve();
    return 0;
  }
  