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;
  long double pi = 3.14159265358979323846;
  void __print(int x) { cerr << x; }
  void __print(long x) { cerr << x; }
  void __print(long long x) { cerr << x; }
  void __print(unsigned x) { cerr << x; }
  void __print(unsigned long x) { cerr << x; }
  void __print(unsigned long long x) { cerr << x; }
  void __print(float x) { cerr << x; }
  void __print(double x) { cerr << x; }
  void __print(long double x) { cerr << x; }
  void __print(char x) { cerr << '\'' << x << '\''; }
  void __print(const char *x) { cerr << '\"' << x << '\"'; }
  void __print(const string &x) { cerr << '\"' << x << '\"'; }
  void __print(bool x) { cerr << (x ? "true" : "false"); }
  template <typename T, typename V>
  void __print(const pair<T, V> &x) {
    cerr << '{';
    __print(x.first);
    cerr << ',';
    __print(x.second);
    cerr << '}';
  }
  template <typename T>
  void __print(const T &x) {
    int f = 0;
    cerr << '{';
    for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
    cerr << "}";
  }
  void _print() { cerr << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    __print(t);
    if (sizeof...(v)) cerr << ", ";
    _print(v...);
  }
  long long int p = 1e9 + 7, n, k;
  long long int dis[101][101];
  void dfs(long long int node, long long int par, vector<long long int> v[]) {
    dis[node][0] = 1;
    for (auto it : v[node]) {
      if (it != par) {
        dfs(it, node, v);
        for (long long int i = 1; i < n; i++) dis[node][i] += dis[it][i - 1];
      }
    }
  }
  long long int cal(vector<long long int> &temp) {
    long long int size = temp.size(), ans = 0, i, j;
    if (size < k) return 0;
    long long int dp[size + 1][k + 1];
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    for (i = 1; i <= size; i++) {
      dp[i][0] = 1;
      for (long long int j = 1; j <= k; j++) {
        dp[i][j] = (dp[i - 1][j] + dp[i - 1][j - 1] * temp[i - 1]) % p;
      }
    }
    return dp[size][k];
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      long long int count;
      cin >> n >> k;
      count = n - 1;
      vector<long long int> v[n];
      while (count--) {
        long long int x, y;
        cin >> x >> y;
        x--, y--;
        v[x].push_back(y);
        v[y].push_back(x);
      }
      if (k == 2) {
        cout << (n * (n - 1)) / 2 << "\n";
        continue;
      }
      long long int ans = 0;
      for (long long int i = 0; i < n; i++) {
        queue<long long int> q;
        q.push(i);
        for (long long int x = 0; x < n; x++)
          for (long long int y = 0; y < n; y++) dis[x][y] = 0;
        dfs(i, -1, v);
        for (long long int x = 0; x < n; x++) {
          vector<long long int> temp;
          for (auto it : v[i]) temp.push_back(dis[it][x]);
          ans = (ans + cal(temp)) % p;
        }
      }
      cout << ans << "\n";
    }
  }
  