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;
  int MOD = 1000000007;
  int expmod(long long int a, long long int b, int m) {
    int res = 1;
    a = a % m;
    if (a == 0) return 0;
    while (b > 0) {
      if (b & 1) {
        res = (res * a) % m;
      }
      b = b >> 1;
      a = (a * a) % m;
    }
    return res;
  }
  long long prodsum(vector<int>& arr, int k) {
    int n = arr.size();
    if (k == 0) return 1;
    if (n < k) return 0;
    if (arr[0] == arr.back()) {
      int common = arr[0];
      long long ans = expmod(common, k, MOD);
      for (int i = 1; i <= k; i++) {
        ans *= n + 1 - i;
        ans %= MOD;
        ans *= expmod(i, MOD - 2, MOD);
        ans %= MOD;
      }
      if (ans < 0) ans += MOD;
      return ans;
    }
    vector<int> part(arr.begin() + 1, arr.end());
    return prodsum(part, k) + arr[0] * prodsum(part, k - 1);
  }
  int solve(vector<vector<int>>& adj, int k) {
    int n = adj.size();
    if (k == 2) {
      return (n * n - n) / 2;
    }
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      set<int> visited = {i};
      vector<map<int, int>> counts;
      for (int nbr : adj[i]) {
        visited.insert(nbr);
        map<int, int> m;
        set<int> current = {nbr};
        int dist = 1;
        m[1] = 1;
        while (current.size()) {
          m[dist] = current.size();
          set<int> next;
          for (int vtx : current) {
            for (int next_nbr : adj[vtx]) {
              if (!visited.count(next_nbr)) {
                next.insert(next_nbr);
                visited.insert(next_nbr);
              }
            }
          }
          current = next;
          dist++;
        }
        counts.push_back(m);
      }
      bool done = false;
      int dist = 1;
      while (!done) {
        vector<int> vals;
        for (map<int, int> count : counts) {
          if (count.find(dist) != count.end()) {
            vals.push_back(count[dist]);
          }
        }
        sort(vals.begin(), vals.end(), greater<int>());
        if (vals.size() < k) {
          done = true;
        } else {
          ans += prodsum(vals, k);
          ans %= MOD;
        }
        dist++;
      }
    }
    return ans;
  }
  int main() {
    int T;
    cin >> T;
    while (T--) {
      int n, k;
      cin >> n >> k;
      vector<vector<int>> adj(n);
      while (n-- > 1) {
        int a, b;
        cin >> a >> b;
        a--, b--;
        adj[a].push_back(b);
        adj[b].push_back(a);
      }
      cout << solve(adj, k) << endl;
    }
  }
  