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 long long mod = 1000000007;
  inline long long read() {
    char ch = getchar();
    long long x = 0, r = 1;
    while (ch < '0' || ch > '9') {
      if (ch == '-') r = 0;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
      x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return r ? x : -x;
  }
  long long t, n, k, x, y, cnt, ans, c, head[101], maxd;
  long long d[101], num[101][101];
  struct node {
    long long v, next;
  } e[201];
  void add(long long u, long long v) {
    e[++cnt].v = v;
    e[cnt].next = head[u];
    head[u] = cnt;
  }
  void dfs(long long x, long long fa, long long color) {
    d[x] = d[fa] + 1;
    maxd = max(maxd, d[x]);
    num[d[x]][color]++;
    for (long long i = head[x]; i; i = e[i].next) {
      long long v = e[i].v;
      if (v == fa) continue;
      dfs(v, x, color);
    }
  }
  long long dp[101][101];
  void update(long long x) {
    d[x] = 1;
    c = 0;
    maxd = 1;
    for (long long i = head[x]; i; i = e[i].next) dfs(e[i].v, x, ++c);
    for (long long i = 1; i <= maxd; ++i) {
      dp[0][0] = 1;
      for (long long j = 1; j <= c; ++j) {
        dp[j][0] = 1;
        for (long long kk = 1; kk <= j; ++kk)
          dp[j][kk] =
              (dp[j - 1][kk] + (dp[j - 1][kk - 1] * num[i][j] % mod)) % mod;
      }
      ans = (ans + dp[c][k]) % mod;
      for (long long j = 1; j <= c; ++j)
        for (long long kk = 0; kk <= j; ++kk) dp[j][kk] = 0;
    }
    for (long long i = 1; i <= maxd; ++i)
      for (long long j = 1; j <= c; ++j) num[i][j] = 0;
  }
  signed main() {
    t = read();
    while (t--) {
      cnt = 0;
      ans = 0;
      memset(head, 0, sizeof(head));
      n = read();
      k = read();
      for (long long i = 1; i < n; ++i) {
        x = read();
        y = read();
        add(x, y);
        add(y, x);
      }
      if (k == 2)
        ans = n * (n - 1) / 2;
      else
        for (long long i = 1; i <= n; ++i) update(i);
      printf("%lld\n", ans);
    }
  }
  