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 long long int MOD = 1e9 + 7;
  std::vector<long long int> v[104];
  long long int mark[104];
  void dfs(long long int in, long long int d,
           std::map<long long int, long long int> &m) {
    long long int i, j;
    mark[in] = 1;
    m[d]++;
    d++;
    for (auto x : v[in]) {
      if (!mark[x]) {
        dfs(x, d, m);
      }
    }
  }
  long long int solve(std::vector<long long int> v, long long int k) {
    long long int i, j, n = ((long long int)v.size());
    if (n == 0) return 0;
    long long int dp[n + 2][k + 2];
    memset(dp, 0, sizeof(dp));
    for (i = 0; i < n; i++) {
      for (j = 1; j < k + 1; j++) {
        if (i == 0) {
          if (j == 1) dp[i][j] += v[i];
          break;
        }
        if (j == 1) {
          dp[i][j] += v[i] + dp[i - 1][j];
          dp[i][j] %= MOD;
        } else {
          dp[i][j] += v[i] * dp[i - 1][j - 1] % MOD;
          dp[i][j] += dp[i - 1][j];
          dp[i][j] %= MOD;
        }
      }
    }
    return dp[n - 1][k];
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int test = 1;
    cin >> test;
    while (test--) {
      long long int n, k, i, j;
      cin >> n >> k;
      for (i = 0; i < n + 1; i++) v[i].clear();
      for (i = 0; i < n - 1; i++) {
        long long int a, b;
        cin >> a >> b;
        v[a].push_back(b);
        v[b].push_back(a);
      }
      if (k == 2) {
        cout << n * (n - 1) / 2 << "\n";
        continue;
      }
      long long int ans = 0;
      for (i = 1; i < n + 1; i++) {
        for (j = 1; j < n + 1; j++) mark[j] = 0;
        mark[i] = 1;
        std::vector<long long int> t[n + 1];
        for (auto x : v[i]) {
          std::map<long long int, long long int> m;
          dfs(x, 1, m);
          for (auto y : m) {
            t[y.first].push_back(y.second);
          }
        }
        for (j = 1; j < n + 1; j++) {
          ans += solve(t[j], k);
          ans %= MOD;
        }
      }
      cout << ans << "\n";
    }
    cerr << "Time Taken:" << (float)clock() / CLOCKS_PER_SEC * 1000 << "ms"
         << "\n";
    return 0;
  }
  