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 md = 1e9 + 7;
  int t, n, k;
  long long cnt[111][111], res[111][111], ans;
  vector<int> v[111];
  long long ft[111], iv[111];
  long long pw(long long x, long long y) {
    long long rt = 1;
    while (y) {
      if (y & 1) {
        rt *= x;
        rt %= md;
      }
      x *= x;
      x %= md;
      y >>= 1;
    }
    return rt;
  }
  long long C(long long x, long long y) {
    return ft[x] * iv[y] % md * iv[x - y] % md;
  }
  void dfs(int x, int y, int rt, int d) {
    cnt[x][d]++;
    for (int i = 0; i < v[x].size(); i++) {
      int z = v[x][i];
      if (z == y) {
        continue;
      }
      dfs(z, x, rt, d + 1);
      if (x != rt) {
        for (int j = 1; j <= n; j++) {
          cnt[x][j] += cnt[z][j];
        }
      } else {
        for (int j = 1; j <= n; j++) {
          for (int kk = k; kk >= 1; kk--) {
            res[j][kk] += res[j][kk - 1] * cnt[z][j] % md;
            res[j][kk] %= md;
          }
        }
      }
    }
  }
  int main() {
    ft[0] = 1;
    iv[0] = 1;
    for (int i = 1; i < 105; i++) {
      ft[i] = ft[i - 1] * i % md;
      iv[i] = pw(ft[i], md - 2);
    }
    cin >> t;
    while (t--) {
      cin >> n >> k;
      for (int i = 1; i <= n; i++) {
        v[i].clear();
      }
      for (int i = 0; i < n - 1; i++) {
        int a, b;
        cin >> a >> b;
        v[a].push_back(b);
        v[b].push_back(a);
      }
      if (k == 2) {
        cout << C(n, 2) << endl;
        continue;
      }
      ans = 0;
      for (int i = 1; i <= n; i++) {
        memset(res, 0, sizeof(res));
        for (int j = 1; j <= n; j++) {
          res[j][0] = 1;
        }
        memset(cnt, 0, sizeof(cnt));
        dfs(i, -1, i, 0);
        for (int j = 1; j <= n; j++) {
          ans += res[j][k];
          ans %= md;
        }
      }
      cout << ans << endl;
    }
    return 0;
  }
  