#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long dp[2000];
int binpow(long long base, long long pow) {
  if (pow == 0) return 1;
  long long half = binpow(base, pow / 2);
  long long ret = (half * half) % mod;
  if (pow % 2 == 1) ret *= base;
  return ret % mod;
}
long long choose(long long a, long long b) {
  if (a < 0 || b < 0 || a < b) return 0;
  long long ret = dp[a];
  long long denom = (dp[a - b] * dp[b]) % mod;
  ret *= binpow(denom, mod - 2);
  ret %= mod;
  return ret;
}
int main() {
  dp[0] = 1;
  for (int i = 1; i < 2000; i++) {
    dp[i] = (dp[i - 1] * i) % mod;
  }
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int T, N, K, u, v;
  cin >> T;
  while (T--) {
    cin >> N >> K;
    long long ret = 0;
    vector<int> adj[N + 1];
    for (int i = 0; i < N - 1; i++) {
      cin >> u >> v;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    for (int i = 1; i <= N; i++) {
      if (adj[i].size() < K) continue;
      vector<pair<long long, int> > bfs;
      vector<int> vis;
      vis.resize(N + 1);
      vis[i] = 1;
      for (int n : adj[i]) {
        bfs.push_back(pair<long long, int>(n, n));
      }
      while (!bfs.empty()) {
        unordered_map<int, int> sz;
        vector<pair<long long, int> > step;
        for (pair<long long, int> j : bfs) {
          sz[j.second]++;
          vis[j.first] = 1;
          for (int n : adj[j.first]) {
            if (!vis[n]) step.push_back(pair<long long, int>(n, j.second));
          }
        }
        if (sz.size() < K) {
          break;
        }
        int second = sz.size();
        vector<int> all;
        for (auto i : sz) {
          all.push_back(i.second);
        }
        long long dp[second + 1][K + 1];
        for (int i = 0; i <= second; i++) {
          for (int j = 0; j <= K; j++) {
            dp[i][j] = (j == 0);
          }
        }
        for (int i = 1; i <= second; i++) {
          for (int j = 1; j <= K; j++) {
            dp[i][j] = dp[i - 1][j];
            if (j != 0) dp[i][j] += dp[i - 1][j - 1] * all[i - 1];
            dp[i][j] %= mod;
          }
        }
        ret += dp[second][K];
        ret %= mod;
        swap(step, bfs);
      }
    }
    cout << (K == 2 ? (N) * (N - 1) / 2 : ret) << '\n';
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long dp[2000];
int binpow(long long base, long long pow) {
  if (pow == 0) return 1;
  long long half = binpow(base, pow / 2);
  long long ret = (half * half) % mod;
  if (pow % 2 == 1) ret *= base;
  return ret % mod;
}
long long choose(long long a, long long b) {
  if (a < 0 || b < 0 || a < b) return 0;
  long long ret = dp[a];
  long long denom = (dp[a - b] * dp[b]) % mod;
  ret *= binpow(denom, mod - 2);
  ret %= mod;
  return ret;
}
int main() {
  dp[0] = 1;
  for (int i = 1; i < 2000; i++) {
    dp[i] = (dp[i - 1] * i) % mod;
  }
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int T, N, K, u, v;
  cin >> T;
  while (T--) {
    cin >> N >> K;
    long long ret = 0;
    vector<int> adj[N + 1];
    for (int i = 0; i < N - 1; i++) {
      cin >> u >> v;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    for (int i = 1; i <= N; i++) {
      if (adj[i].size() < K) continue;
      vector<pair<long long, int> > bfs;
      vector<int> vis;
      vis.resize(N + 1);
      vis[i] = 1;
      for (int n : adj[i]) {
        bfs.push_back(pair<long long, int>(n, n));
      }
      while (!bfs.empty()) {
        unordered_map<int, int> sz;
        vector<pair<long long, int> > step;
        for (pair<long long, int> j : bfs) {
          sz[j.second]++;
          vis[j.first] = 1;
          for (int n : adj[j.first]) {
            if (!vis[n]) step.push_back(pair<long long, int>(n, j.second));
          }
        }
        if (sz.size() < K) {
          break;
        }
        int second = sz.size();
        vector<int> all;
        for (auto i : sz) {
          all.push_back(i.second);
        }
        long long dp[second + 1][K + 1];
        for (int i = 0; i <= second; i++) {
          for (int j = 0; j <= K; j++) {
            dp[i][j] = (j == 0);
          }
        }
        for (int i = 1; i <= second; i++) {
          for (int j = 1; j <= K; j++) {
            dp[i][j] = dp[i - 1][j];
            if (j != 0) dp[i][j] += dp[i - 1][j - 1] * all[i - 1];
            dp[i][j] %= mod;
          }
        }
        ret += dp[second][K];
        ret %= mod;
        swap(step, bfs);
      }
    }
    cout << (K == 2 ? (N) * (N - 1) / 2 : ret) << '\n';
  }
  return 0;
}

```