#include <bits/stdc++.h>
using namespace std;
using pii = pair<long long, long long>;
const long long INF = 1e9;
const long long MOD = 1e9 + 7;
const long long N = 105;
long long d[N];
void dfs(long long node, long long par, long long depth,
         vector<vector<long long>>& graph) {
  ++d[depth];
  for (auto i : graph[node]) {
    if (i == par) continue;
    dfs(i, node, depth + 1, graph);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    vector<vector<long long>> graph(n + 1);
    for (long long i = 0; i < n - 1; i++) {
      long long u, v;
      cin >> u >> v;
      graph[u].emplace_back(v);
      graph[v].emplace_back(u);
    }
    if (k == 2) {
      long long ans = (n * (n - 1) / 2) % MOD;
      cout << ans << "\n";
      continue;
    }
    long long ans = 0;
    for (long long root = 1; root <= n; root++) {
      vector<vector<long long>> deps(n + 1);
      for (auto c : graph[root]) {
        memset(d, 0, sizeof d);
        dfs(c, root, 1, graph);
        for (long long i = 1; i <= n; i++) {
          if (d[i]) deps[i].emplace_back(d[i]);
        }
      }
      for (auto dep : deps) {
        long long sz = dep.size();
        if (sz < k) continue;
        vector<vector<long long>> dp(sz + 1, vector<long long>(k + 1));
        dp[0][0] = 1;
        for (long long i = 1; i <= sz; i++) {
          for (long long j = 0; j <= k; j++) {
            dp[i][j] = dp[i - 1][j];
            if (j)
              dp[i][j] = (dp[i][j] + (dp[i - 1][j - 1] * dep[i - 1])) % MOD;
          }
        }
        ans = (ans + dp[sz][k]) % MOD;
      }
    }
    cout << ans << "\n";
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
using pii = pair<long long, long long>;
const long long INF = 1e9;
const long long MOD = 1e9 + 7;
const long long N = 105;
long long d[N];
void dfs(long long node, long long par, long long depth,
         vector<vector<long long>>& graph) {
  ++d[depth];
  for (auto i : graph[node]) {
    if (i == par) continue;
    dfs(i, node, depth + 1, graph);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    vector<vector<long long>> graph(n + 1);
    for (long long i = 0; i < n - 1; i++) {
      long long u, v;
      cin >> u >> v;
      graph[u].emplace_back(v);
      graph[v].emplace_back(u);
    }
    if (k == 2) {
      long long ans = (n * (n - 1) / 2) % MOD;
      cout << ans << "\n";
      continue;
    }
    long long ans = 0;
    for (long long root = 1; root <= n; root++) {
      vector<vector<long long>> deps(n + 1);
      for (auto c : graph[root]) {
        memset(d, 0, sizeof d);
        dfs(c, root, 1, graph);
        for (long long i = 1; i <= n; i++) {
          if (d[i]) deps[i].emplace_back(d[i]);
        }
      }
      for (auto dep : deps) {
        long long sz = dep.size();
        if (sz < k) continue;
        vector<vector<long long>> dp(sz + 1, vector<long long>(k + 1));
        dp[0][0] = 1;
        for (long long i = 1; i <= sz; i++) {
          for (long long j = 0; j <= k; j++) {
            dp[i][j] = dp[i - 1][j];
            if (j)
              dp[i][j] = (dp[i][j] + (dp[i - 1][j - 1] * dep[i - 1])) % MOD;
          }
        }
        ans = (ans + dp[sz][k]) % MOD;
      }
    }
    cout << ans << "\n";
  }
}

```