#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int t;
int n, k;
vector<int> g[110];
int dep[110], dp[110][110];
inline void upd(int &x, int y) { x = (x + y) % mod; }
void dfs(int x, int fa, int d, int &cnt) {
  if (dep[x] == d) cnt++;
  for (int i = 0; i < (int)g[x].size(); i++) {
    int to = g[x][i];
    if (to == fa) continue;
    dep[to] = dep[x] + 1;
    dfs(to, x, d, cnt);
  }
}
void solve() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) g[i].clear();
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  if (k == 2) {
    cout << 1ll * n * (n - 1) / 2 % mod << endl;
    return;
  }
  int ans = 0;
  for (int r = 0; r < n; r++)
    for (int d = 1; d <= n; d++) {
      if ((int)g[r].size() < k) continue;
      dp[0][0] = 1;
      for (int i = 0; i < (int)g[r].size(); i++) {
        int cnt = 0;
        dep[g[r][i]] = 1;
        dfs(g[r][i], r, d, cnt);
        for (int j = 0; j <= i + 1; j++) dp[i + 1][j] = 0;
        for (int j = 0; j <= i; j++) {
          if (cnt > 0) upd(dp[i + 1][j + 1], 1ll * dp[i][j] * cnt % mod);
          upd(dp[i + 1][j], dp[i][j]);
        }
      }
      cout << endl;
      upd(ans, dp[(int)g[r].size()][k]);
    }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### 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;
const int mod = 1e9 + 7;
int t;
int n, k;
vector<int> g[110];
int dep[110], dp[110][110];
inline void upd(int &x, int y) { x = (x + y) % mod; }
void dfs(int x, int fa, int d, int &cnt) {
  if (dep[x] == d) cnt++;
  for (int i = 0; i < (int)g[x].size(); i++) {
    int to = g[x][i];
    if (to == fa) continue;
    dep[to] = dep[x] + 1;
    dfs(to, x, d, cnt);
  }
}
void solve() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) g[i].clear();
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  if (k == 2) {
    cout << 1ll * n * (n - 1) / 2 % mod << endl;
    return;
  }
  int ans = 0;
  for (int r = 0; r < n; r++)
    for (int d = 1; d <= n; d++) {
      if ((int)g[r].size() < k) continue;
      dp[0][0] = 1;
      for (int i = 0; i < (int)g[r].size(); i++) {
        int cnt = 0;
        dep[g[r][i]] = 1;
        dfs(g[r][i], r, d, cnt);
        for (int j = 0; j <= i + 1; j++) dp[i + 1][j] = 0;
        for (int j = 0; j <= i; j++) {
          if (cnt > 0) upd(dp[i + 1][j + 1], 1ll * dp[i][j] * cnt % mod);
          upd(dp[i + 1][j], dp[i][j]);
        }
      }
      cout << endl;
      upd(ans, dp[(int)g[r].size()][k]);
    }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```