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 maxn = 110;
  const long long mod = 1e9 + 7;
  void speed() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
  }
  int n, k, used[maxn];
  long long dp[maxn][maxn];
  vector<int> g[maxn];
  void bfs(int v) {
    queue<int> q;
    for (int i = 1; i <= n; i++) used[i] = -1;
    used[v] = 0;
    q.push(v);
    while (!q.empty()) {
      v = q.front();
      q.pop();
      for (int i = 0; i < g[v].size(); i++) {
        int u = g[v][i];
        if (used[u] == -1) {
          used[u] = used[v] + 1;
          q.push(u);
        }
      }
    }
  }
  long long dfs(int v, int par, int dis) {
    if (used[v] == dis) return 1;
    long long sum = 0;
    for (int j = 0; j < g[v].size(); j++) {
      int u = g[v][j];
      if (u == par) continue;
      sum = sum + dfs(u, v, dis);
    }
    return sum;
  }
  void solve() {
    cin >> n >> k;
    for (int i = 1; i <= n; i++) g[i].clear();
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      g[v].push_back(u);
      g[u].push_back(v);
    }
    if (k > 2) {
      long long ans = 0;
      for (int dis = 2; dis <= n; dis += 2) {
        for (int i = 1; i <= n; i++) {
          bfs(i);
          dp[0][0] = 1;
          int id = 1;
          for (int j = 0; j < g[i].size(); j++) {
            long long sum = dfs(g[i][j], i, dis / 2);
            dp[id][0] = 1;
            for (int p = 1; p <= k; p++)
              dp[id][p] = (dp[id - 1][p] + dp[id - 1][p - 1] * sum) % mod;
            id++;
          }
          ans = (ans + dp[id - 1][k]) % mod;
        }
      }
      cout << ans << '\n';
    } else {
      long long ans = 0;
      for (int dis = 1; dis <= n; dis++) {
        for (int i = 1; i <= n; i++) {
          bfs(i);
          for (int j = 1; j <= n; j++)
            if (used[j] == dis) ans++;
        }
      }
      cout << ans / 2 << '\n';
    }
  }
  int main() {
    int t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  