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 N = 105;
  const int p = 1e9 + 7;
  int read() {
    int s = 0;
    char c = getchar(), lc = '+';
    while (c < '0' || '9' < c) lc = c, c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
    return lc == '-' ? -s : s;
  }
  void write(int x) {
    if (x < 0) putchar('-'), x = -x;
    if (x < 10)
      putchar(x + '0');
    else
      write(x / 10), putchar(x % 10 + '0');
  }
  void print(int x, char c = '\n') {
    write(x);
    putchar(c);
  }
  struct edge {
    int to, nxt;
  } e[N];
  int head[N], cnte = 0;
  void add_edge(int u, int v) {
    e[++cnte].to = v;
    e[cnte].nxt = head[u];
    head[u] = cnte;
  }
  void clear() {
    cnte = 0;
    memset(head, 0, sizeof(head));
  }
  vector<int> a[N];
  int dp[N], deep[N], tot[N];
  int dfs(int now, int father) {
    tot[deep[now] = deep[father] + 1]++;
    int ret = deep[now];
    for (int i = head[now]; i; i = e[i].nxt) {
      int to = e[i].to;
      if (to == father) continue;
      ret = max(ret, dfs(to, now));
    }
    return ret;
  }
  signed main(signed Goodbye, char *Wangang[]) {
    (void)Goodbye, (void)Wangang;
    int T = read();
    while (T--) {
      int n = read(), k = read(), ans = 0;
      for (int i = 1; i < n; i++) {
        int u = read(), v = read();
        add_edge(u, v);
        add_edge(v, u);
      }
      if (k == 2) {
        print(n * (n - 1) / 2), clear();
        continue;
      }
      for (int rt = 1; rt <= n; rt++) {
        deep[rt] = 0;
        for (int i = head[rt]; i; i = e[i].nxt) {
          int to = e[i].to, maxdeep = dfs(to, rt);
          for (int j = 1; j <= maxdeep; j++) {
            a[j].push_back(tot[j]);
            tot[j] = 0;
          }
        }
        for (int i = 1; i <= n; i++)
          if (!a[i].empty()) {
            memset(dp, 0, sizeof(dp));
            dp[0] = 1;
            for (int x : a[i])
              for (int j = n; j >= 1; j--)
                dp[j] = (dp[j] + 1LL * dp[j - 1] * x) % p;
            ans = (ans + dp[k]) % p;
            a[i].clear();
          }
      }
      print(ans);
      clear();
    }
    return 0;
  }
  