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 maxn = 2e5 + 7;
  long long mod = 1e9 + 7;
  int n, k;
  vector<int> e[maxn];
  int f[102][102], num[102][102];
  void dfs(int u, int p, int len, int fa) {
    num[fa][len]++;
    for (int v : e[u]) {
      if (v == p) continue;
      dfs(v, u, len + 1, fa);
    }
  }
  int cal(vector<int> v) {
    memset(f, 0, sizeof(f));
    int res = 0;
    int m = v.size();
    f[0][0] = 1;
    for (int i = 1; i <= m; i++) {
      for (int j = 0; j <= i; j++) {
        if (j)
          f[i][j] = (f[i - 1][j] + 1ll * f[i - 1][j - 1] * v[i - 1]) % mod;
        else
          f[i][j] = f[i - 1][j];
      }
    }
    res = f[m][k] % mod;
    return res;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int _;
    cin >> _;
    while (_--) {
      cin >> n >> k;
      for (int i = (1); i <= (n - 1); ++i) {
        int u, v;
        cin >> u >> v;
        e[u].push_back(v), e[v].push_back(u);
      }
      if (k == 2) {
        cout << (n * (n - 1) / 2ll) % mod << endl;
        for (int i = (1); i <= (n); ++i) e[i].clear();
        continue;
      }
      int ans = 0;
      for (int i = 1; i <= n; i++) {
        memset(num, 0, sizeof(num));
        for (int v : e[i]) dfs(v, i, 1, v);
        for (int d = 1; d <= n; d++) {
          vector<int> vv;
          for (int v : e[i]) vv.push_back(num[v][d]);
          ans = (ans + cal(vv)) % mod;
        }
      }
      cout << ans << endl;
      for (int i = (1); i <= (n); ++i) e[i].clear();
    }
    return 0;
  }
  