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 nax = 125, mod = 1e9 + 7;
  long long add(long long x, long long y) {
    x += y;
    if (x >= mod) x -= mod;
    return x;
  }
  long long mul(long long x, long long y) { return x * y % mod; }
  vector<long long> g[nax];
  long long vis[nax], cnt[nax], dp[nax][nax];
  long long find(long long n, long long k) {
    for (long long i = 0; i <= n; i++)
      for (long long j = 0; j <= k; j++) dp[i][j] = 0;
    dp[0][0] = 1;
    for (long long i = 0; i < n; i++)
      for (long long j = 0; j <= k; j++) {
        dp[i + 1][j] = add(dp[i + 1][j], dp[i][j]);
        dp[i + 1][j + 1] = add(dp[i + 1][j + 1], mul(dp[i][j], cnt[i]));
      }
    return dp[n][k];
  }
  int32_t main() {
    long long t;
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      for (long long i = 0; i < n; i++) g[i].clear();
      for (long long i = 1; i < n; i++) {
        long long x, y;
        cin >> x >> y;
        g[--x].push_back(--y);
        g[y].push_back(x);
      }
      if (k == 2) {
        cout << (n * (n - 1) / 2) % mod << '\n';
        continue;
      }
      long long res = 0;
      for (long long x = 0; x < n; x++) {
        memset(vis, 0, sizeof(vis));
        vis[x] = 1;
        vector<pair<long long, long long> > e;
        long long m = g[x].size();
        for (long long i = 0; i < m; i++) {
          long long y = g[x][i];
          e.push_back({y, i});
          cnt[i] = 1;
          vis[y] = 1;
        }
        while (!e.empty()) {
          res = add(res, find(m, k));
          vector<pair<long long, long long> > ee;
          for (auto p : e) {
            cnt[p.second]--;
            for (auto y : g[p.first])
              if (!vis[y]) {
                ee.push_back({y, p.second});
                vis[y] = 1;
                cnt[p.second]++;
              }
          }
          e = ee;
        }
      }
      cout << res << '\n';
    }
    return 0;
  }
  