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>
  #pragma GCC optimize("Ofast")
  #pragma GCC optimize("unroll-loops")
  #pragma GCC target("sse,sse2,sse3,ssse3,abm,mmx,tune=native")
  using namespace std;
  long long gcd(long long i, long long j) {
    if (j == 0)
      return i;
    else
      return gcd(j, i % j);
  }
  template <typename T>
  inline T getint() {
    T val = 0;
    char c;
    bool neg = false;
    while ((c = getchar()) && !(c >= '0' && c <= '9')) {
      neg |= c == '-';
    }
    do {
      val = (val * 10) + c - '0';
    } while ((c = getchar()) && (c >= '0' && c <= '9'));
    return val * (neg ? -1 : 1);
  }
  const long long INF = 1e18 + 100;
  const int mod = 1000000007;
  const long double eps = 1e-10, pi = acosl(-1);
  const long long maxN = 200010, maxT = 25000, A = 179, mid = 150;
  mt19937 mt_rand(time(0));
  long long bp(long long et, long long b) {
    b %= mod - 1;
    long long res = 1;
    for (int i = 30; i >= 0; --i) {
      res = (res * res) % mod;
      if ((b & (1 << i)) != 0) res = (res * et) % mod;
    }
    return res;
  }
  void panic() {
    cout << "-1\n";
    exit(0);
  }
  vector<vector<int>> g;
  vector<int> cnt;
  void dfs(int v, int d, int p = -1) {
    ++cnt[d];
    for (auto x : g[v]) {
      if (x == p) continue;
      dfs(x, d + 1, v);
    }
  }
  void solve() {
    int n, k;
    cin >> n >> k;
    g.assign(n, vector<int>());
    for (int i = 0; i < n - 1; ++i) {
      int a, b;
      cin >> a >> b;
      --a, --b;
      g[a].push_back(b);
      g[b].push_back(a);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << "\n";
      return;
    }
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      vector<vector<int>> dp(n, vector<int>(k + 1));
      for (int i = 0; i < n; ++i) dp[i][0] = 1;
      for (auto x : g[i]) {
        cnt.assign(n, 0);
        dfs(x, 1, i);
        for (int d = 0; d < n; ++d) {
          if (cnt[d] == 0) continue;
          for (int j = k - 1; j >= 0; --j) {
            dp[d][j + 1] = (dp[d][j + 1] + (long long)cnt[d] * dp[d][j]) % mod;
          }
        }
      }
      for (int i = 0; i < n; ++i) {
        ans = (ans + dp[i][k]) % mod;
      }
    }
    cout << ans << "\n";
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cout.precision(10);
    int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  