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;
  long long n, k, dis[102][102], ans, tmm, mod = 1e9 + 7, vis[102][102],
                                           dp[102][102];
  vector<int> gr[102];
  vector<long long> v;
  void dfs(int x, int p) {
    dis[x][0]++;
    for (int i = 0; i < gr[x].size(); i++) {
      int u = gr[x][i];
      if (u == p) continue;
      dfs(u, x);
      for (int j = 0; j <= n; j++) dis[x][j] += dis[u][j - 1];
    }
  }
  long long f(int x, int nm) {
    if (nm == k) return 1;
    if (x == v.size()) return 0;
    if (vis[x][nm] == tmm) return dp[x][nm];
    long long ans = 0;
    vis[x][nm] = tmm;
    ans = (ans + f(x + 1, nm)) % mod;
    ans = (ans + f(x + 1, nm + 1) * v[x]) % mod;
    return dp[x][nm] = ans;
  }
  void root(int x, int p) {
    for (int j = 1; j <= n; j++) {
      for (int z = 0; z < gr[x].size(); z++) {
        int u = gr[x][z];
        if (dis[u][j - 1] > 0) v.push_back(dis[u][j - 1]);
      }
      if (v.size() >= k) {
        tmm++;
        ans = (ans + f(0, 0)) % mod;
      }
      v.clear();
    }
    for (int i = 0; i < gr[x].size(); i++) {
      int u = gr[x][i];
      if (u == p) continue;
      for (int j = 1; j <= n; j++) dis[x][j] -= dis[u][j - 1];
      for (int j = 1; j <= n; j++) dis[u][j] += dis[x][j - 1];
      root(u, x);
      for (int j = 1; j <= n; j++) dis[u][j] -= dis[x][j - 1];
      for (int j = 1; j <= n; j++) dis[x][j] += dis[u][j - 1];
    }
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      ans = 0;
      cin >> n >> k;
      for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= 100; j++) dis[i][j] = 0;
        gr[i].clear();
      }
      for (int i = 0; i < n - 1; i++) {
        int x, y;
        cin >> x >> y;
        gr[x].push_back(y);
        gr[y].push_back(x);
      }
      if (k == 2) {
        cout << (n * (n - 1)) / 2 << endl;
        continue;
      }
      dfs(1, 0);
      root(1, 0);
      cout << ans << endl;
    }
  }
  