#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
vector<vector<long long> > d(110);
vector<bool> u(110);
vector<long long> t;
long long dfs(long long v, long long dist, long long pr) {
  u[v] = 1;
  if (!dist) {
    return 1;
  }
  long long m = 0;
  for (long long i = 0; i < d[v].size(); i++)
    if (!u[d[v][i]]) {
      long long b = dfs(d[v][i], dist - 1, pr);
      m += b;
      if (v == pr && b) t.push_back(b);
    }
  return m;
}
long long f(long long k) {
  long long m = mod - 2;
  long long a = 1;
  while (m) {
    if (m % 2) a = 1ll * a * k % mod;
    k = 1ll * k * k % mod;
    m /= 2;
  }
  return a;
}
void solve() {
  long long n, m, k, ans = 0;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) d[i].clear();
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    u--;
    v--;
    d[u].push_back(v);
    d[v].push_back(u);
  }
  if (k == 1) {
    cout << n << '\n';
    return;
  }
  if (k == 2) {
    cout << n * (n - 1) / 2 << '\n';
    return;
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 1; j < n; j++) {
      long long ans1 = 0;
      for (long long h = 0; h < n; h++) u[h] = 0;
      t.clear();
      m = dfs(i, j, i);
      if (t.size() >= k) {
        long long dp[110][110] = {0};
        for (long long j = 0; j < k; j++) {
          long long sum;
          if (!j)
            sum = 1;
          else
            sum = dp[j - 1][j - 1];
          for (long long i = j; i < t.size(); i++) {
            dp[i][j] = 1ll * sum * t[i] % mod;
            if (j) sum = (sum + dp[i][j - 1]) % mod;
          }
        }
        for (long long i = 0; i < t.size(); i++)
          ans1 = (ans1 + dp[i][k - 1]) % mod;
        ans = (ans + ans1) % mod;
      }
    }
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Please create a solution in CPP 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;
long long mod = 1e9 + 7;
vector<vector<long long> > d(110);
vector<bool> u(110);
vector<long long> t;
long long dfs(long long v, long long dist, long long pr) {
  u[v] = 1;
  if (!dist) {
    return 1;
  }
  long long m = 0;
  for (long long i = 0; i < d[v].size(); i++)
    if (!u[d[v][i]]) {
      long long b = dfs(d[v][i], dist - 1, pr);
      m += b;
      if (v == pr && b) t.push_back(b);
    }
  return m;
}
long long f(long long k) {
  long long m = mod - 2;
  long long a = 1;
  while (m) {
    if (m % 2) a = 1ll * a * k % mod;
    k = 1ll * k * k % mod;
    m /= 2;
  }
  return a;
}
void solve() {
  long long n, m, k, ans = 0;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) d[i].clear();
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    u--;
    v--;
    d[u].push_back(v);
    d[v].push_back(u);
  }
  if (k == 1) {
    cout << n << '\n';
    return;
  }
  if (k == 2) {
    cout << n * (n - 1) / 2 << '\n';
    return;
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 1; j < n; j++) {
      long long ans1 = 0;
      for (long long h = 0; h < n; h++) u[h] = 0;
      t.clear();
      m = dfs(i, j, i);
      if (t.size() >= k) {
        long long dp[110][110] = {0};
        for (long long j = 0; j < k; j++) {
          long long sum;
          if (!j)
            sum = 1;
          else
            sum = dp[j - 1][j - 1];
          for (long long i = j; i < t.size(); i++) {
            dp[i][j] = 1ll * sum * t[i] % mod;
            if (j) sum = (sum + dp[i][j - 1]) % mod;
          }
        }
        for (long long i = 0; i < t.size(); i++)
          ans1 = (ans1 + dp[i][k - 1]) % mod;
        ans = (ans + ans1) % mod;
      }
    }
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```