#include <bits/stdc++.h>
using namespace std;
int d[1000005];
int f[505][505];
vector<int> g[1000005];
void dfs(int u, int r, int pr, int val, int dis = 1) {
  if (dis == val) d[r]++;
  for (auto &v : g[u])
    if (v != pr) dfs(v, r, u, val, dis + 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t-- > 0) {
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; ++i) g[i].clear();
    for (int i = 1; i < n; ++i) {
      int u, v;
      cin >> u >> v;
      g[u].emplace_back(v);
      g[v].emplace_back(u);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << endl;
      continue;
    }
    int res = 0;
    for (int i = 1; i <= n; ++i) {
      if ((g[i]).size() < k) continue;
      for (int j = 1; j < n; ++j) {
        for (auto &v : g[i]) d[v] = 0;
        for (auto &v : g[i]) dfs(v, v, i, j);
        for (int first = 0; first <= (g[i]).size(); ++first)
          for (int second = 0; second <= k; ++second) f[first][second] = 0;
        f[0][0] = 1;
        for (int first = 1; first <= (g[i]).size(); ++first) {
          for (int second = 0; second <= k; ++second) {
            f[first][second] = f[first - 1][second];
            if (second > 0)
              f[first][second] =
                  (f[first][second] + (long long)f[first - 1][second - 1] *
                                          d[g[i][first - 1]] % 1000000007) %
                  1000000007;
          }
        }
        res = (res + f[(g[i]).size()][k]) % 1000000007;
      }
    }
    cout << res << endl;
  }
}

### Prompt

Please formulate a Cpp solution to the following problem:

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;
int d[1000005];
int f[505][505];
vector<int> g[1000005];
void dfs(int u, int r, int pr, int val, int dis = 1) {
  if (dis == val) d[r]++;
  for (auto &v : g[u])
    if (v != pr) dfs(v, r, u, val, dis + 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t-- > 0) {
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; ++i) g[i].clear();
    for (int i = 1; i < n; ++i) {
      int u, v;
      cin >> u >> v;
      g[u].emplace_back(v);
      g[v].emplace_back(u);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << endl;
      continue;
    }
    int res = 0;
    for (int i = 1; i <= n; ++i) {
      if ((g[i]).size() < k) continue;
      for (int j = 1; j < n; ++j) {
        for (auto &v : g[i]) d[v] = 0;
        for (auto &v : g[i]) dfs(v, v, i, j);
        for (int first = 0; first <= (g[i]).size(); ++first)
          for (int second = 0; second <= k; ++second) f[first][second] = 0;
        f[0][0] = 1;
        for (int first = 1; first <= (g[i]).size(); ++first) {
          for (int second = 0; second <= k; ++second) {
            f[first][second] = f[first - 1][second];
            if (second > 0)
              f[first][second] =
                  (f[first][second] + (long long)f[first - 1][second - 1] *
                                          d[g[i][first - 1]] % 1000000007) %
                  1000000007;
          }
        }
        res = (res + f[(g[i]).size()][k]) % 1000000007;
      }
    }
    cout << res << endl;
  }
}

```