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;
  void PV(vector<long long> v) {
    for (long long i = 0; i < (long long)v.size(); i++) cout << v[i] << " ";
    cout << "\n";
  }
  void PVV(vector<pair<long long, long long> > v) {
    for (long long i = 0; i < (long long)v.size(); i++)
      cout << v[i].first << " " << v[i].second << "\n";
  }
  void PA(long long v[], long long n, long long x = 0) {
    for (long long i = x; i < n + x; i++) cout << v[i] << ' ';
    cout << "\n";
  }
  void IN(long long a[], long long n, long long x = 0) {
    for (long long i = x; i < n + x; i++) cin >> a[i];
  }
  inline void op() {}
  const long long M = 1e9 + 7;
  const long long IM = 1e18 + 37;
  const long long N = 1000 + 10;
  vector<long long> v[N];
  map<long long, long long> cn[N];
  long long dp[N][N];
  long long mx = 0;
  long long n, k;
  void dfs(long long x, long long baap, long long col, long long l) {
    mx = max(mx, l);
    cn[col][l]++;
    for (long long it : v[x]) {
      if (it != baap) dfs(it, x, col, l + 1);
    }
  }
  long long solve(long long l, long long k, long long d) {
    if (l == 0) return (k == 0);
    if (dp[l][k] != -1) return dp[l][k];
    return dp[l][k] =
               (solve(l - 1, k, d) + (solve(l - 1, k - 1, d) * cn[l][d]) % M) % M;
  }
  void clear(long long n) {
    for (long long i = 0; i <= n; i++) fill(dp[i], dp[i] + k + 10, -1);
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    op();
    long long t;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      for (long long i = 0; i <= n; i++) {
        v[i].clear();
      }
      for (long long i = 1; i < n; i++) {
        long long x, y;
        cin >> x >> y;
        v[x].push_back(y);
        v[y].push_back(x);
      }
      long long ans = 0, col;
      if (k == 2) {
        ans = (n * (n - 1) / 2) % M;
        cout << (ans) << "\n";
        continue;
      }
      for (long long i = 1; i <= n; i++) {
        mx = 0, col = 1;
        for (long long it : v[i]) dfs(it, i, col++, 1);
        for (long long dis = 1; dis <= mx; dis++) {
          clear(col);
          ans = (ans + solve(col, k, dis)) % M;
        }
        for (long long j = 1; j <= col; j++) cn[j].clear();
      }
      cout << (ans) << "\n";
    }
    return 0;
  }
  