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 mod = 1000000007LL;
  long long C[105][105];
  int dis[105], headers[105];
  int cnt;
  struct line {
    int from;
    int to;
    int next;
  };
  struct line que[205];
  void add(int from, int to) {
    cnt++;
    que[cnt].from = from;
    que[cnt].to = to;
    que[cnt].next = headers[from];
    headers[from] = cnt;
  }
  int depth[105];
  void dfs(int place, int father) {
    dis[place] = dis[father] + 1;
    depth[dis[place]]++;
    for (int i = headers[place]; i; i = que[i].next)
      if (que[i].to != father) dfs(que[i].to, place);
  }
  int main() {
    int n, t, u, v, k;
    scanf("%d", &t);
    for (int i = 0; i <= 100; i++) C[i][0] = 1;
    for (int i = 1; i <= 100; i++)
      for (int j = 1; j <= i; j++)
        C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
    while (t--) {
      scanf("%d%d", &n, &k);
      cnt = 0;
      for (int i = 1; i <= n; i++) headers[i] = 0;
      for (int i = 1; i < n; i++) {
        scanf("%d%d", &u, &v);
        add(u, v);
        add(v, u);
      }
      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>> son;
        for (int j = headers[i]; j; j = que[j].next) {
          vector<int> temp;
          temp.push_back(0);
          for (int k = 1; k <= n; k++) depth[k] = 0;
          dis[i] = 0;
          dfs(que[j].to, i);
          for (int j = 1; j <= n; j++) temp.push_back(depth[j]);
          son.push_back(temp);
        }
        vector<vector<long long>> f(son.size() + 1, vector<long long>(k + 1, 0));
        for (int i = 0; i <= son.size(); i++) f[i][0] = 1;
        for (int o = 1; o <= n; o++) {
          for (int j = 1; j <= son.size(); j++)
            for (int l = 1; l <= k; l++)
              f[j][l] = (f[j - 1][l] + f[j - 1][l - 1] * son[j - 1][o]) % mod;
          ans = (ans + f[son.size()][k]) % mod;
        }
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  