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 int mod = 1e9 + 7;
  int head[maxn], sons[maxn][maxn];
  long long int c[maxn][maxn], dp[maxn][maxn];
  vector<int> son;
  struct edge {
    int to, next;
  } e[maxn * 2];
  int cnt, maxdep, mindep, k;
  void add(int x, int y) {
    e[cnt].to = y;
    e[cnt].next = head[x];
    head[x] = cnt++;
  }
  void dfs(int rt, int u, int fa, int deep) {
    sons[rt][deep]++;
    for (int i = head[u]; i != -1; i = e[i].next) {
      int v = e[i].to;
      if (v == fa) continue;
      dfs(rt, v, u, deep + 1);
    }
  }
  long long int cal(vector<int> son) {
    memset(dp, 0, sizeof(dp));
    int n = son.size() - 1;
    dp[0][0] = 1;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j <= i; j++) {
        if (j) {
          dp[i][j] += dp[i - 1][j - 1] * son[i];
        }
        dp[i][j] += dp[i - 1][j];
        dp[i][j] %= mod;
      }
    }
    return dp[n][k];
  }
  int main() {
    c[0][0] = 1;
    for (int i = 1; i < maxn; i++) {
      c[i][0] = 1;
      for (int j = 1; j <= i; j++) {
        c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
      }
    }
    int t, n, x, y;
    cin >> t;
    while (t--) {
      long long int ans = 0;
      cnt = 0;
      memset(head, -1, sizeof(head));
      cin >> n >> k;
      for (int i = 1; i < n; i++) {
        cin >> x >> y;
        add(x, y);
        add(y, x);
      }
      if (k == 2) {
        cout << c[n][2] << endl;
        continue;
      }
      for (int i = 1; i <= n; i++) {
        memset(sons, 0, sizeof(sons));
        for (int u = head[i]; u != -1; u = e[u].next) {
          int v = e[u].to;
          dfs(v, v, i, 1);
        }
        for (int j = 1; j <= 100; j++) {
          long long int tmp = 1;
          int tot = 0;
          son.clear();
          son.push_back(-1);
          for (int u = head[i]; u != -1; u = e[u].next) {
            int v = e[u].to;
            son.push_back(sons[v][j]);
          }
          ans = (ans + cal(son)) % mod;
        }
      }
      cout << ans << endl;
    }
    return 0;
  }
  