#include <bits/stdc++.h>
using namespace std;
const long long N = 105, MOD = 1e9 + 7;
vector<vector<long long>> adj;
map<long long, map<long long, long long>> cnt;
long long dep[N], ans, dp[N];
long long n, k;
void dfs(long long u, long long p, long long root) {
  cnt[root][dep[u]]++;
  for (auto v : adj[u]) {
    if (v == p) continue;
    dep[v] = dep[u] + 1;
    dfs(v, u, root);
  }
}
void solve() {
  cin >> n >> k;
  adj.clear();
  adj.resize(n + 1);
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  ans = 0;
  if (k == 2) {
    ans = n * (n - 1) / 2;
  } else {
    for (long long i = 1; i <= n; i++) {
      cnt.clear();
      for (auto v : adj[i]) {
        dep[v] = 1;
        dfs(v, i, v);
      }
      for (long long j = 1; j <= 100; j++) {
        fill(dp, dp + N, 0);
        dp[0] = 1;
        for (auto v : adj[i]) {
          for (long long kk = 100; ~kk; kk--) {
            dp[kk + 1] = (dp[kk + 1] + dp[kk] * cnt[v][j] % MOD);
          }
        }
        ans = (ans + dp[k]) % 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();
}

### Prompt

Generate 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 long long N = 105, MOD = 1e9 + 7;
vector<vector<long long>> adj;
map<long long, map<long long, long long>> cnt;
long long dep[N], ans, dp[N];
long long n, k;
void dfs(long long u, long long p, long long root) {
  cnt[root][dep[u]]++;
  for (auto v : adj[u]) {
    if (v == p) continue;
    dep[v] = dep[u] + 1;
    dfs(v, u, root);
  }
}
void solve() {
  cin >> n >> k;
  adj.clear();
  adj.resize(n + 1);
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  ans = 0;
  if (k == 2) {
    ans = n * (n - 1) / 2;
  } else {
    for (long long i = 1; i <= n; i++) {
      cnt.clear();
      for (auto v : adj[i]) {
        dep[v] = 1;
        dfs(v, i, v);
      }
      for (long long j = 1; j <= 100; j++) {
        fill(dp, dp + N, 0);
        dp[0] = 1;
        for (auto v : adj[i]) {
          for (long long kk = 100; ~kk; kk--) {
            dp[kk + 1] = (dp[kk + 1] + dp[kk] * cnt[v][j] % MOD);
          }
        }
        ans = (ans + dp[k]) % 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();
}

```