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>
  const long long inf = 0x3f3f3f3f;
  const double dinf = 1e100;
  using namespace std;
  const long long maxn = 205;
  const long long mod = 1e9 + 7;
  long long n, k;
  long long d[maxn];
  long long a[maxn][maxn];
  vector<long long> e[maxn];
  long long dp[maxn][maxn], w[maxn];
  void dfs(long long u, long long fa, long long rt) {
    d[u] = d[fa] + 1, a[rt][d[u]]++;
    for (auto x : e[u]) {
      if (x == fa) continue;
      dfs(x, u, rt);
    }
  }
  signed main() {
    ios::sync_with_stdio(false), cin.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      for (long long i = (1); i <= (n); ++i) e[i].clear();
      for (long long i = (1); i <= ((n - 1)); ++i) {
        long long l, r;
        cin >> l >> r;
        e[l].push_back(r), e[r].push_back(l);
      }
      if (k == 2) {
        cout << n * (n - 1) / 2 << endl;
      } else {
        long long ans = 0;
        for (long long i = (1); i <= (n); ++i) {
          d[i] = 0;
          for (long long j = 0; j < e[i].size(); ++j) {
            dfs(e[i][j], i, e[i][j]);
          }
          for (long long j = (1); j <= (n); ++j) {
            long long p = 0;
            for (auto x : e[i]) {
              w[++p] = a[x][j];
            }
            memset(dp, 0, sizeof(dp));
            dp[0][0] = 1;
            for (long long i = (1); i <= (p); ++i) {
              for (long long j = 0; j <= min(i, k); ++j) {
                if (j > 0)
                  dp[i][j] = (dp[i][j] + dp[i - 1][j - 1] * w[i] % mod) % mod;
                dp[i][j] = (dp[i][j] + dp[i - 1][j]) % mod;
              }
            }
            ans = (ans + dp[p][k]) % mod;
          }
          memset(a, 0, sizeof(a));
        }
        cout << ans << endl;
      }
    }
    return 0;
  }
  