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 MOD = 1e9 + 7;
  const int N = 1e2 + 5;
  int add(int a, int b) {
    int res = a + b;
    if (res >= MOD) res -= MOD;
    return res;
  }
  int mult(int a, int b) {
    long long res = 1LL * a * b;
    if (res >= MOD) res %= MOD;
    return res;
  }
  int ans;
  vector<vector<int>> g;
  int dfs(int node, int p, int dist) {
    if (dist == 0) return 1;
    int res = 0;
    for (auto to : g[node]) {
      if (to != p) res += dfs(to, node, dist - 1);
    }
    return res;
  }
  vector<int> delta;
  vector<vector<int>> dp;
  int max_n;
  int solve(int p, int left) {
    if (left == 0) return 1;
    if (p >= max_n) return 0;
    int &res = dp[p][left];
    if (~res) return res;
    res = 0;
    res = add(res, solve(p + 1, left));
    res = add(res, mult(delta[p], solve(p + 1, left - 1)));
    return res;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int tt;
    cin >> tt;
    while (tt--) {
      int n, k;
      cin >> n >> k;
      ans = 0;
      g.clear();
      g.resize(n);
      for (int i = 0; i < n - 1; i++) {
        int st, et;
        cin >> st >> et;
        --st;
        --et;
        g[st].emplace_back(et);
        g[et].emplace_back(st);
      }
      if (k == 2) {
        for (int i = n - 1; i >= 1; i--) {
          ans = add(ans, i);
        }
      } else {
        for (int distance = 1; distance <= n; distance++) {
          for (int i = 0; i < n; i++) {
            delta.clear();
            for (auto to : g[i]) delta.emplace_back(dfs(to, i, distance - 1));
            max_n = (int)delta.size();
            if (max_n < k) continue;
            dp = vector<vector<int>>(max_n, vector<int>(k + 1, -1));
            ans = add(ans, solve(0, k));
          }
        }
      }
      cout << ans << '\n';
    }
    return 0;
  }
  