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;
  long long nCrModp(int n, int r, int p) {
    if (r > n - r) r = n - r;
    long long C[r + 1];
    memset(C, 0, sizeof(C));
    C[0] = 1;
    for (int i = 1; i <= n; i++) {
      for (int j = min(i, r); j > 0; j--) C[j] = (C[j] + C[j - 1]) % p;
    }
    return C[r];
  }
  int n, k;
  vector<vector<int> > graph(105);
  vector<bool> visited(105);
  long long NCR[105][105];
  int dfs(int curr, int dist) {
    visited[curr] = true;
    if (dist == 0) return 1;
    int ans = 0;
    for (int a = 0; a < graph[curr].size(); a++) {
      int x = graph[curr][a];
      if (visited[x]) continue;
      ans += dfs(x, dist - 1);
    }
    return ans;
  }
  void solve() {
    cin >> n >> k;
    for (int a = 0; a <= n; a++) {
      graph[a].clear();
    }
    int x, y;
    for (int a = 1; a < n; a++) {
      cin >> x >> y;
      graph[x].push_back(y);
      graph[y].push_back(x);
    }
    if (k == 2) {
      cout << NCR[n][2] << '\n';
      return;
    }
    long long ans = 0;
    long long pairs = 0;
    long long dp[k + 1][2];
    for (int a = 2; a < n; a += 2) {
      for (int b = 1; b <= n; b++) {
        if (graph[b].size() < k) {
          continue;
        }
        for (int c = 1; c <= n; c++) {
          visited[c] = false;
        }
        for (int c = 0; c <= k; c++) {
          dp[c][0] = 0;
          dp[c][1] = 0;
        }
        dp[0][1] = 1;
        dp[0][0] = 1;
        visited[b] = true;
        int cnt = 0, last, now;
        for (int c = 0; c < graph[b].size(); c++) {
          now = c % 2;
          last = (c + 1) % 2;
          for (int d = 0; d <= min(c + 1, k); d++) {
            dp[d][now] = dp[d][last];
          }
          cnt = dfs(graph[b][c], a / 2 - 1);
          if (cnt == 0) {
            continue;
          }
          for (int d = 1; d <= min(c + 1, k); d++) {
            dp[d][now] += (cnt * dp[d - 1][last]) % 1000000007;
            dp[d][now] %= 1000000007;
          }
        }
        ans += dp[k][now];
        ans %= 1000000007;
      }
    }
    cout << ans << '\n';
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    for (int a = 0; a < 105; a++) {
      for (int b = 0; b <= a; b++) {
        NCR[a][b] = nCrModp(a, b, 1000000007);
      }
    }
    int tt = 1;
    cin >> tt;
    while (tt--) {
      solve();
    }
    return 0;
  }
  