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;
  const int MOD = 1e9 + 7;
  int modadd(int x, int y) {
    int result = x + y;
    if (result >= MOD) result -= MOD;
    return result;
  }
  int modmult(int x, int y) { return (x * int64_t(y)) % MOD; }
  int modaddto(int& x, int y) {
    x += y;
    if (x >= MOD) x -= MOD;
    return x;
  }
  int64_t modpow(int64_t b, int64_t e) {
    int64_t ans = 1;
    for (; e; b = b * b % MOD, e /= 2)
      if (e & 1) ans = ans * b % MOD;
    return ans;
  }
  int modinv(int x) { return modpow(x, MOD - 2); }
  vector<int> get_parents(const vector<vector<int>>& adj, int root) {
    int n = adj.size();
    vector<int> parents(n, -1);
    stack<pair<int, int>> st;
    st.emplace(root, -1);
    while (!st.empty()) {
      int node = st.top().first;
      parents[node] = st.top().second;
      st.pop();
      for (int i : adj[node]) {
        if (i != parents[node]) {
          st.emplace(i, node);
        }
      }
    }
    return parents;
  }
  vector<vector<int>> get_children(const vector<int>& parents) {
    int n = parents.size();
    vector<vector<int>> children(n);
    for (int i = 0; i < n; ++i) {
      if (parents[i] >= 0) {
        children[parents[i]].push_back(i);
      }
    }
    return children;
  }
  vector<int> get_depth(const vector<vector<int>>& children, int root) {
    int n = children.size();
    vector<int> depth(n);
    vector<int> q(1, root);
    int d = 0;
    while (!q.empty()) {
      vector<int> q2;
      for (int i : q) {
        depth[i] = d;
        for (int j : children[i]) {
          q2.push_back(j);
        }
      }
      q2.swap(q);
      d++;
    }
    return depth;
  }
  int solve(const vector<int>& u, const vector<int>& v, int k) {
    int n = u.size() + 1;
    if (k == 2) {
      return n * (n - 1) / 2;
    }
    vector<vector<int>> adj(n);
    for (int i = 0; i < n - 1; ++i) {
      adj[u[i]].push_back(v[i]);
      adj[v[i]].push_back(u[i]);
    }
    int ways = 0;
    for (int root = 0; root < n; ++root) {
      if (int(adj[root].size()) < k) continue;
      vector<int> parents = get_parents(adj, root);
      vector<vector<int>> children = get_children(parents);
      vector<int> depth = get_depth(children, root);
      int max_depth = *max_element(depth.begin(), depth.end());
      int nch = children[root].size();
      vector<vector<int>> depths(nch, vector<int>(max_depth, 0));
      for (int chi = 0; chi < nch; ++chi) {
        stack<int> st;
        st.push(children[root][chi]);
        while (!st.empty()) {
          int node = st.top();
          st.pop();
          depths[chi][depth[node] - 1]++;
          for (int j : children[node]) {
            st.push(j);
          }
        }
      }
      for (int d = 0; d < max_depth; ++d) {
        vector<int> dp(1, 1);
        for (int chi = 0; chi < nch; ++chi) {
          vector<int> dp2(chi + 2, 0);
          dp2[0] = dp[0];
          for (int i = 1; i < chi + 2; ++i) {
            dp2[i] = (i < chi + 1) ? dp[i] : 0;
            modaddto(dp2[i], modmult(dp[i - 1], depths[chi][d]));
          }
          dp2.swap(dp);
        }
        assert(int(dp.size()) > k);
        modaddto(ways, dp[k]);
      }
    }
    return ways;
  }
  void test_case() {
    int n, k;
    cin >> n >> k;
    vector<int> u(n - 1), v(n - 1);
    for (int i = 0; i < n - 1; ++i) {
      cin >> u[i] >> v[i];
      --u[i];
      --v[i];
    }
    cout << solve(u, v, k) << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    for (int i = 0; i < t; ++i) {
      test_case();
    }
  }
  