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 M = 205;
  const int MOD = 1e9 + 7;
  inline int read() {
    char c = getchar();
    int x = 0, f = 1;
    for (; c < '0' || c > '9'; c = getchar())
      if (c == '-') f = 0;
    for (; c <= '9' && c >= '0'; c = getchar())
      x = (x << 1) + (x << 3) + (c ^ 48);
    return f ? x : -x;
  }
  inline int max(int a, int b) { return a > b ? a : b; }
  inline int min(int a, int b) { return a < b ? a : b; }
  int n, m, num[M][M], f[M][M];
  vector<int> a;
  struct edge {
    int v, nxt;
  } ed[M << 1];
  int head[M], cnt_edge;
  inline void add_edge(int u, int v) {
    ed[++cnt_edge] = (edge){v, head[u]}, head[u] = cnt_edge;
  }
  inline void dfs(int x, int fa) {
    num[x][0] = 1;
    for (int i = head[x]; i; i = ed[i].nxt) {
      int v = ed[i].v;
      if (v != fa) {
        dfs(v, x);
        for (int j = 1; j <= n; j++) num[x][j] += num[v][j - 1];
      }
    }
  }
  int main() {
    int T = read();
    while (T--) {
      n = read(), m = read();
      int ans = 0;
      cnt_edge = 0;
      for (int i = 0; i <= n; i++) head[i] = 0;
      for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= n; j++) num[i][j] = 0;
      }
      for (int i = 1; i < n; i++) {
        int u = read(), v = read();
        add_edge(u, v), add_edge(v, u);
      }
      if (m == 2) {
        printf("%d\n", n * (n - 1) / 2);
        continue;
      }
      for (int i = 1; i <= n; i++) {
        dfs(i, 0);
        for (int j = 1; j <= n; j++) {
          for (int k = 1; k <= a.size(); k++) {
            for (int l = 1; l <= k; l++) f[k][l] = 0;
          }
          a.clear();
          for (int l = head[i]; l; l = ed[l].nxt)
            a.push_back(num[ed[l].v][j - 1]);
          f[0][0] = 1;
          for (int k = 1; k <= a.size(); k++) {
            f[k][0] = 1;
            for (int l = 1; l <= k; l++)
              f[k][l] = (f[k - 1][l] + f[k - 1][l - 1] * a[k - 1] % MOD) % MOD;
          }
          ans = (ans + f[a.size()][m]) % MOD;
        }
        for (int j = 1; j <= n; j++) {
          for (int l = 0; l <= n; l++) num[j][l] = 0;
        }
      }
      printf("%d\n", ans);
    }
    return 0;
  }
  