#include <bits/stdc++.h>
using namespace std;
const int N = 110;
int dp[N][N];
int cnt[N];
int dep[N];
vector<int> v[N];
const int mod = 1e9 + 7;
void dfs(int u, int fa, int d) {
  cnt[d]++;
  for (auto j : v[u]) {
    if (j == fa) continue;
    dfs(j, u, d + 1);
  }
}
void solve() {
  memset(dp, 0, sizeof dp);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) v[i].clear();
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  if (m == 2) {
    cout << n * (n - 1) / 2 << endl;
    return;
  }
  int ans = 0;
  for (int root = 1; root <= n; root++) {
    int ct = 0;
    for (int dd = 1; dd <= n; dd++) {
      dp[root][0] = 1;
      for (auto j : v[root]) {
        for (int kk = 1; kk <= n; kk++) cnt[kk] = 0;
        dfs(j, root, 1);
        if (cnt[dd])
          for (int k = m; k >= 1; k--)
            dp[root][k] =
                (1ll * cnt[dd] * dp[root][k - 1] % mod + dp[root][k]) % mod;
      }
      ans = (ans + dp[root][m]) % mod;
      for (int j = 1; j <= m; j++) dp[root][j] = 0;
    }
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
const int N = 110;
int dp[N][N];
int cnt[N];
int dep[N];
vector<int> v[N];
const int mod = 1e9 + 7;
void dfs(int u, int fa, int d) {
  cnt[d]++;
  for (auto j : v[u]) {
    if (j == fa) continue;
    dfs(j, u, d + 1);
  }
}
void solve() {
  memset(dp, 0, sizeof dp);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) v[i].clear();
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  if (m == 2) {
    cout << n * (n - 1) / 2 << endl;
    return;
  }
  int ans = 0;
  for (int root = 1; root <= n; root++) {
    int ct = 0;
    for (int dd = 1; dd <= n; dd++) {
      dp[root][0] = 1;
      for (auto j : v[root]) {
        for (int kk = 1; kk <= n; kk++) cnt[kk] = 0;
        dfs(j, root, 1);
        if (cnt[dd])
          for (int k = m; k >= 1; k--)
            dp[root][k] =
                (1ll * cnt[dd] * dp[root][k - 1] % mod + dp[root][k]) % mod;
      }
      ans = (ans + dp[root][m]) % mod;
      for (int j = 1; j <= m; j++) dp[root][j] = 0;
    }
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```