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 N = 105, INF = 1 << 30, MOD = 1e9 + 7;
  int n, k;
  vector<int> adj[N];
  long long cnt[N][N];
  long long ans = 0, F[N];
  long long sumOfProducts(vector<int>& arr, int n, int k) {
    long long dp[n + 1][n + 1];
    memset(dp, 0, sizeof(dp));
    long long cur_sum = 0;
    for (int i = 1; i <= n; i++) {
      dp[1][i] = arr[i - 1];
      (cur_sum += arr[i - 1]) %= MOD;
    }
    for (int i = 2; i <= k; ++i) {
      long long temp_sum = 0;
      for (int j = 1; j <= n; j++) {
        (cur_sum -= dp[i - 1][j]) %= MOD;
        dp[i][j] = (arr[j - 1] * cur_sum) % MOD;
        (temp_sum += dp[i][j]) %= MOD;
      }
      cur_sum = temp_sum;
    }
    return (cur_sum + MOD) % MOD;
  }
  void dfs(int u, int root, int p, int d) {
    ++cnt[root][d];
    for (int v : adj[u]) {
      if (v == p) continue;
      dfs(v, root, u, d + 1);
    }
  }
  void solve(int u) {
    memset(cnt, 0, sizeof(cnt));
    if (adj[u].size() < k) return;
    for (int v : adj[u]) {
      dfs(v, v, u, 0);
    }
    for (int i = 0; i < n; ++i) {
      int CNT = 0;
      for (int v : adj[u])
        if (cnt[v][i]) ++CNT;
      if (CNT < k) break;
      vector<int> C;
      for (int v : adj[u])
        if (cnt[v][i]) C.push_back(cnt[v][i]);
      (ans += sumOfProducts(C, CNT, k)) %= MOD;
    }
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    F[0] = 1;
    for (int i = 1; i < N; ++i) F[i] = (F[i - 1] * i) % MOD;
    int t;
    cin >> t;
    while (t--) {
      ans = 0;
      cin >> n >> k;
      for (int i = 0; i < n; ++i) adj[i].clear();
      for (int i = 0; i < n - 1; ++i) {
        int u, v;
        cin >> u >> v;
        adj[--u].push_back(--v);
        adj[v].push_back(u);
      }
      if (k == 2)
        cout << n * (n - 1) / 2 << '\n';
      else {
        for (int i = 0; i < n; ++i) solve(i);
        cout << ans % MOD << '\n';
      }
    }
  }
  