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>
  #pragma GCC optimize("Ofast")
  using namespace std;
  const long long N = 1e3;
  const long long MOD = 1e9 + 7;
  const long long inf = 1e10;
  const long double eps = 1e-12;
  const long double EPS = 1e-10;
  long long rx[] = {1, -1, 0, 0};
  long long ry[] = {0, 0, 1, -1};
  vector<long long> g[N];
  long long used[N], dp[N][N], cnt[N];
  long long rundp(long long m, long long k) {
    for (long long i = 0; i <= m; ++i) {
      for (long long j = 0; j <= k; ++j) {
        dp[i][j] = 0;
      }
    }
    dp[0][0] = 1;
    for (long long i = 0; i < m; ++i) {
      for (long long j = 0; j <= k; ++j) {
        dp[i + 1][j] += dp[i][j];
        dp[i + 1][j] %= MOD;
        dp[i + 1][j + 1] += (dp[i][j] * cnt[i]);
        dp[i + 1][j + 1] %= MOD;
      }
    }
    return dp[m][k];
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long Q;
    cin >> Q;
    while (Q--) {
      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;
        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 ans = 0;
      for (long long c = 0; c < n; ++c) {
        fill(used, used + n, 0);
        used[c] = 1;
        vector<pair<long long, long long> > layer;
        long long m = g[c].size();
        for (long long i = 0; i < m; ++i) {
          long long x = g[c][i];
          layer.push_back({x, i});
          cnt[i] = 1;
          used[x] = 1;
        }
        while (!layer.empty()) {
          ans = ans + rundp(m, k);
          ans %= MOD;
          vector<pair<long long, long long> > nlayer;
          for (auto to : layer) {
            cnt[to.second]--;
            for (auto x : g[to.first]) {
              if (used[x]) continue;
              nlayer.push_back({x, to.second});
              used[x] = 1;
              cnt[to.second]++;
            }
          }
          layer = nlayer;
        }
      }
      cout << ans << '\n';
    }
    return 0;
  }
  