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;
  using LL = long long;
  const LL mod = 1e9 + 7;
  int n, k, u, v;
  vector<int> graph[205];
  LL dp[205][205];
  bool vis[205];
  LL solve(int m, int k, vector<int> &cnt) {
    dp[0][0] = 1;
    for (int j = 1; j <= m; ++j) {
      for (int l = 0; l <= k; ++l) {
        dp[j][l] = (dp[j - 1][l] + dp[j][l]) % mod;
        if (l > 0) {
          dp[j][l] = (dp[j - 1][l - 1] * cnt[j - 1] % mod + dp[j][l]) % mod;
        }
      }
    }
    LL ans = dp[m][k];
    for (int j = 0; j <= m; ++j) {
      for (int l = 0; l <= k; ++l) {
        dp[j][l] = 0;
      }
    }
    fill_n(cnt.begin(), m, 0);
    return ans;
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
      scanf("%d%d", &n, &k);
      for (int i = 1; i < n; ++i) {
        scanf("%d%d", &u, &v);
        graph[u].push_back(v);
        graph[v].push_back(u);
      }
      if (k > 2) {
        LL ans = 0;
        for (int i = 1; i <= n; ++i) {
          fill_n(vis + 1, n, false);
          vis[i] = true;
          vector<int> cnt(graph[i].size(), 0);
          queue<tuple<int, int, int>> quu;
          for (int j = 0; j < graph[i].size(); ++j) {
            quu.push({graph[i][j], j, 1});
          }
          int nowHeight = 1;
          while (!quu.empty()) {
            auto cur = quu.front();
            quu.pop();
            vis[get<0>(cur)] = true;
            if (get<2>(cur) > nowHeight) {
              ans = (ans + solve(graph[i].size(), k, cnt)) % mod;
              ++nowHeight;
            }
            ++cnt[get<1>(cur)];
            for (auto nxt : graph[get<0>(cur)]) {
              if (vis[nxt]) continue;
              quu.push({nxt, get<1>(cur), get<2>(cur) + 1});
            }
          }
          ans = (ans + solve(graph[i].size(), k, cnt)) % mod;
        }
        printf("%lld\n", ans);
      } else {
        printf("%lld\n", 1ll * n * (n - 1) / 2 % mod);
      }
      for (int i = 1; i <= n; ++i) {
        graph[i].clear();
      }
    }
    return 0;
  }
  