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>
  const int INF = 2147483647;
  const int INF2 = 0x3f3f3f3f;
  const long long INF64 = (long long)1e18;
  const double INFD = 1e30;
  const double EPS = 1e-9;
  const double PI = std::acos(-1);
  const int MOD = 1e9 + 7;
  template <typename T>
  inline T read() {
    T X = 0, w = 0;
    char ch = 0;
    while (!isdigit(ch)) {
      w |= ch == '-';
      ch = getchar();
    }
    while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
  }
  const int MAXN = 105;
  const int MAXV = (1 << 17) + 5;
  int dr[4] = {0, 1, 0, -1};
  int dc[4] = {-1, 0, 1, 0};
  int dr2[8] = {1, 1, 1, -1, -1, -1, 0, 0};
  int dc2[8] = {1, 0, -1, 1, 0, -1, 1, -1};
  int CASE = 1;
  int A[MAXN];
  int fact[MAXN];
  std::vector<int> G[MAXN];
  int dep[MAXN];
  int cnt[MAXN][MAXN];
  int dp[MAXN][MAXN];
  int fastExp(int x, int p, int mod) {
    int ans = 1;
    while (p) {
      if (p & 1) ans = (long long)ans * x % mod;
      x = (long long)x * x % mod;
      p >>= 1;
    }
    return ans;
  }
  int getInv(int x, int mod) { return fastExp(x, MOD - 2, MOD); }
  int choose(int n, int k) {
    if (k > n || n < 0 || k < 0) return 0;
    return fact[n] * getInv(fact[k] * fact[n - k] % MOD, MOD) % MOD;
  }
  void dfs(int x, int p) {
    for (auto v : G[x]) {
      if (v == p) continue;
      dep[v] = dep[x] + 1;
      dfs(v, x);
    }
  }
  void dfscnt(int x, int p, int pp) {
    cnt[pp][dep[x]]++;
    for (auto v : G[x]) {
      if (v == p) continue;
      dfscnt(v, x, pp);
    }
  }
  void solve() {
    int n, k;
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) G[i].clear();
    for (int i = 1; i <= n - 1; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      G[u].push_back(v);
      G[v].push_back(u);
    }
    if (k == 2) {
      printf("%d\n", ((long long)n * (n - 1) / 2) % MOD);
      return;
    }
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      dep[i] = 0;
      dfs(i, i);
      int sz = G[i].size();
      if (sz < k) continue;
      for (int a = 0; a <= n; a++) {
        for (int b = 0; b <= n; b++) {
          cnt[a][b] = 0;
        }
      }
      for (int j = 0; j < sz; j++) {
        auto v = G[i][j];
        dfscnt(v, i, v);
      }
      for (int s = 1; s <= n; s++) {
        dp[0][0] = 1;
        for (int j = 1; j <= sz; j++) {
          auto v = G[i][j - 1];
          for (int l = 0; l <= k; l++) {
            dp[j][l] = dp[j - 1][l];
            if (l > 0) {
              dp[j][l] = dp[j][l] + (long long)cnt[v][s] * dp[j - 1][l - 1] % MOD;
              dp[j][l] %= MOD;
            }
          }
        }
        ans += dp[sz][k];
        ans %= MOD;
      }
    }
    printf("%d\n", ans);
  }
  int main() {
    fact[0] = 1;
    for (int i = 1; i <= 100; i++) {
      fact[i] = (long long)fact[i - 1] * i % MOD;
    }
    int T = read<int>();
    while (T--) {
      solve();
    }
    return 0;
  }
  