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 INF = (is_same<int, long long>::value ? 1e18 + 666 : 1e9 + 666);
  mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
  template <class t1, class t2>
  inline bool cmin(t1 &a, const t2 &b) {
    if (a > b) {
      a = b;
      return true;
    }
    return false;
  }
  template <class t1, class t2>
  inline bool cmax(t1 &a, const t2 &b) {
    if (a < b) {
      a = b;
      return true;
    }
    return false;
  }
  inline int bpow(long long a, long long b, int mod) {
    int res = 1;
    a %= mod;
    for (; b; b >>= 1) {
      if (b & 1) {
        res = int((1ll * a * res) % mod);
        res %= mod;
      }
      a *= a;
      a %= mod;
    }
    return res;
  }
  void UseFiles(const string &s) {
    freopen((s + ".in").c_str(), "r", stdin);
    freopen((s + ".out").c_str(), "w", stdout);
  }
  void run();
  signed main() {
    iostream::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    run();
  }
  void solve();
  void run() {
    int n;
    cin >> n;
    while (n-- > 0) {
      solve();
    }
  }
  void solve() {
    constexpr int mod = 1e9 + 7;
    int n, k;
    cin >> n >> k;
    vector<vector<int>> g(n);
    for (int i = 0; i < n - 1; ++i) {
      int u, v;
      cin >> u >> v;
      --u, --v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    if (k == 2) {
      cout << (n * (n - 1) >> 1) << endl;
      return;
    }
    function<vector<int>(int, int)> dfs = [&g, &dfs](int v, int p) {
      vector<int> res;
      for (auto &x : g[v]) {
        if (x == p) continue;
        vector<int> ret = dfs(x, v);
        if (((int)(ret).size()) > ((int)(res).size())) {
          swap(ret, res);
        }
        for (int i = 1; i <= ((int)(ret).size()); ++i) {
          res[((int)(res).size()) - i] += ret[((int)(ret).size()) - i];
        }
      }
      res.push_back(1);
      return res;
    };
    int ans = 0;
    for (int center = 0; center < n; ++center) {
      if (((int)(g[center]).size()) < k) continue;
      vector<vector<int>> dverts;
      for (auto &v : g[center]) {
        dverts.push_back(dfs(v, center));
        reverse(dverts.back().begin(), dverts.back().end());
      }
      for (int i = 0;; ++i) {
        vector<int> dp(k + 1, 0);
        dp[0] = 1;
        for (auto &x : dverts) {
          if (((int)(x).size()) <= i) continue;
          for (int j = k; j > 0; --j) {
            dp[j] += int(1ll * dp[j - 1] * x[i] % mod);
            if (dp[j] >= mod) {
              dp[j] -= mod;
            }
          }
        }
        if (dp[k] == 0) break;
        ans += dp[k];
        if (ans >= mod) {
          ans -= mod;
        }
      }
    }
    cout << ans << endl;
  }
  