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 INF = 1e9 + 7;
  const int N = 1e2 + 10;
  vector<int> v[N];
  void dfs(int x, vector<int>& cnt, int fa = 0, int d = 0) {
    cnt[d]++;
    for (auto& y : v[x]) {
      if (y == fa) continue;
      dfs(y, cnt, x, d + 1);
    }
  }
  vector<int> multy(vector<int>& v1, vector<int>& v2) {
    while (v1.size() > 1 && v1.back() == 0) v1.pop_back();
    while (v2.size() > 1 && v2.back() == 0) v2.pop_back();
    vector<int> ret(v1.size() + v2.size() - 1);
    for (int i = 0; i < v1.size(); i++) {
      for (int j = 0; j < v2.size(); j++) {
        ret[i + j] = (ret[i + j] + 1LL * v1[i] * v2[j]) % INF;
      }
    }
    return ret;
  }
  int solve(vector<int>& vp, int k) {
    if (vp.size() < k) return 0;
    vector<int> ret;
    ret.push_back(1);
    for (auto& x : vp) {
      vector<int> v{1, x};
      ret = multy(ret, v);
    }
    if (ret.size() <= k)
      return 0;
    else
      return ret[k];
  }
  int main() {
    int ncase;
    scanf("%d", &ncase);
    while (ncase--) {
      int n, k;
      scanf("%d%d", &n, &k);
      for (int i = 1; i <= n; i++) v[i].clear();
      for (int i = 1; i < n; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        v[x].push_back(y);
        v[y].push_back(x);
      }
      if (k == 2) {
        printf("%d\n", n * (n - 1) / 2);
        continue;
      }
      long long ans = 0;
      for (int i = 1; i <= n; i++) {
        vector<vector<int>> key(n);
        for (auto& j : v[i]) {
          vector<int> cnt(n, 0);
          dfs(j, cnt, i, 0);
          for (int k = 0; k < n; k++) {
            if (cnt[k] > 0) key[k].push_back(cnt[k]);
          }
        }
        for (auto& vp : key) {
          ans += solve(vp, k);
        }
      }
      printf("%lld\n", ans % INF);
    }
    return 0;
  }
  