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;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  inline void boostIO() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cout << fixed << setprecision(0);
  }
  bool isPrime(int x) {
    if (x <= 4 || x % 2 == 0 || x % 3 == 0) return x == 2 || x == 3;
    for (int i = 5; i * i <= x; i += 6)
      if (x % i == 0 || x % (i + 2) == 0) return 0;
    return 1;
  }
  long long int gcd(long long int a, long long int b) {
    return b ? gcd(b, a % b) : a;
  }
  long long int lcm(long long int a, long long int b) {
    return a * b / gcd(a, b);
  }
  bool BREAKTIME(long double v) { return 1.0 * clock() >= v * CLOCKS_PER_SEC; }
  long long mod = 1e9 + 7;
  const long long inf = 1e18;
  long long n, k;
  vector<long long> D;
  void dfs(long long a, vector<vector<long long>>& adj, long long p) {
    D[a] = D[p] + 1;
    for (auto b : adj[a]) {
      if (b == p) continue;
      dfs(b, adj, a);
    }
  }
  int32_t main() {
    boostIO();
    long long T;
    cin >> T;
    while (T--) {
      cin >> n >> k;
      vector<vector<long long>> adj(n + 1);
      for (long long i = 0; i < (n - 1); ++i) {
        long long a, b;
        cin >> a >> b;
        adj[a].push_back(b);
        adj[b].push_back(a);
      }
      long long ans = 0;
      long long additional = 0;
      for (long long s = 1; s <= n; ++s) {
        vector<vector<long long>> A(n + 1);
        for (auto a : adj[s]) {
          D = vector<long long>(n + 1);
          dfs(a, adj, s);
          vector<long long> d(n + 1);
          for (long long i = 1; i <= n; ++i) {
            d[D[i]]++;
          }
          for (long long i = 1; i <= n; ++i) {
            if (d[i] != 0) A[i].push_back(d[i]);
          }
        }
        for (auto v : A) {
          if (v.size() == 0) continue;
          vector<long long> e(v.size());
          for (long long i = 0; i < v.size(); ++i) {
            e[i] = v[i];
            if (i != 0) {
              e[i] += e[i - 1];
              e[i] %= mod;
            }
          }
          for (long long j = 0; j < k - 1; ++j) {
            vector<long long> e2(v.size());
            for (long long i = 0; i < v.size(); ++i) {
              if (i != 0) {
                e2[i] = e[i - 1] * v[i];
                e2[i] += e2[i - 1];
                e2[i] %= mod;
              }
            }
            e = e2;
          }
          ans += e.back();
          ans %= mod;
        }
      }
      if (k == 2) {
        ans = n * (n - 1) / 2;
      }
      cout << ans << '\n';
    }
  }
  