#include <bits/stdc++.h>
using namespace std;
using cd = complex<double>;
const double PI = acos(-1);
const int N = 109;
int c[N][N];
int mod = 1e9 + 7;
vector<vector<int>> adj;
int d[N][N];
void calc(int node, int par, int dep, int root) {
  d[root][dep]++;
  for (int ch : adj[node])
    if (ch != par) calc(ch, node, dep + 1, root);
}
void solve() {
  int n, k;
  cin >> n >> k;
  adj = vector<vector<int>>(n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u - 1].push_back(v - 1);
    adj[v - 1].push_back(u - 1);
  }
  if (k == 2) {
    cout << c[n][k] << "\n";
    return;
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    memset(d, 0, sizeof d);
    int cnt = adj[i].size();
    vector<int> v;
    for (int ch : adj[i]) {
      calc(ch, i, 0, ch);
      v.push_back(ch);
    }
    if (cnt < k) continue;
    for (int di = 0; di <= n; di++) {
      int dp[cnt + 1][k + 1];
      memset(dp, 0, sizeof dp);
      dp[0][0] = 1;
      for (int i = 1; i <= cnt; i++) {
        dp[i][0] = 1;
        for (int r = 1; r <= k; r++) {
          dp[i][r] =
              (dp[i - 1][r] + 1ll * dp[i - 1][r - 1] * d[v[i - 1]][di] % mod) %
              mod;
        }
      }
      ans += dp[cnt][k];
      ans %= mod;
    }
  }
  cout << ans << "\n";
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  c[0][0] = 1;
  for (int i = 1; i < N; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
  }
  int t;
  cin >> t;
  while (t--) solve();
}

### 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 cd = complex<double>;
const double PI = acos(-1);
const int N = 109;
int c[N][N];
int mod = 1e9 + 7;
vector<vector<int>> adj;
int d[N][N];
void calc(int node, int par, int dep, int root) {
  d[root][dep]++;
  for (int ch : adj[node])
    if (ch != par) calc(ch, node, dep + 1, root);
}
void solve() {
  int n, k;
  cin >> n >> k;
  adj = vector<vector<int>>(n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u - 1].push_back(v - 1);
    adj[v - 1].push_back(u - 1);
  }
  if (k == 2) {
    cout << c[n][k] << "\n";
    return;
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    memset(d, 0, sizeof d);
    int cnt = adj[i].size();
    vector<int> v;
    for (int ch : adj[i]) {
      calc(ch, i, 0, ch);
      v.push_back(ch);
    }
    if (cnt < k) continue;
    for (int di = 0; di <= n; di++) {
      int dp[cnt + 1][k + 1];
      memset(dp, 0, sizeof dp);
      dp[0][0] = 1;
      for (int i = 1; i <= cnt; i++) {
        dp[i][0] = 1;
        for (int r = 1; r <= k; r++) {
          dp[i][r] =
              (dp[i - 1][r] + 1ll * dp[i - 1][r - 1] * d[v[i - 1]][di] % mod) %
              mod;
        }
      }
      ans += dp[cnt][k];
      ans %= mod;
    }
  }
  cout << ans << "\n";
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  c[0][0] = 1;
  for (int i = 1; i < N; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
  }
  int t;
  cin >> t;
  while (t--) solve();
}

```