#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long maxn = 100 + 5, mod = 1e9 + 7, inf = mod;
long long n, k;
vector<long long> g[maxn];
long long ans;
long long cnt[maxn];
vector<long long> dis[maxn];
long long dp[maxn];
void dfs(long long i, long long p, long long d) {
  cnt[d]++;
  for (long long j : g[i]) {
    if (j != p) {
      dfs(j, i, d + 1);
    }
  }
}
long long get(vector<long long>& v) {
  fill(begin(dp), end(dp), 0);
  dp[0] = 1;
  for (long long x : v) {
    for (long long num = (long long)n; num >= (long long)1; num--) {
      dp[num] = (dp[num] + dp[num - 1] * x) % mod;
    }
  }
  return dp[k];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tc;
  cin >> tc;
  while (tc--) {
    cin >> n >> k;
    ans = 0;
    fill(g + 1, g + n + 1, vector<long long>());
    for (long long i = (long long)1; i <= (long long)n - 1; i++) {
      long long u, v;
      cin >> u >> v;
      g[u].emplace_back(v);
      g[v].emplace_back(u);
    }
    if (k == 1) {
      cout << n << '\n';
      continue;
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << '\n';
      continue;
    }
    for (long long i = (long long)1; i <= (long long)n; i++) {
      for (long long d = (long long)0; d <= (long long)n; d++) dis[d].clear();
      for (long long j : g[i]) {
        fill(begin(cnt), end(cnt), 0);
        dfs(j, i, 1);
        for (long long d = (long long)1; d <= (long long)n; d++) {
          dis[d].emplace_back(cnt[d]);
        }
      }
      for (long long d = (long long)1; d <= (long long)n; d++) {
        ans += get(dis[d]);
        ans %= mod;
      }
    }
    cout << ans << '\n';
  }
}

### Prompt

Develop a solution in CPP 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long maxn = 100 + 5, mod = 1e9 + 7, inf = mod;
long long n, k;
vector<long long> g[maxn];
long long ans;
long long cnt[maxn];
vector<long long> dis[maxn];
long long dp[maxn];
void dfs(long long i, long long p, long long d) {
  cnt[d]++;
  for (long long j : g[i]) {
    if (j != p) {
      dfs(j, i, d + 1);
    }
  }
}
long long get(vector<long long>& v) {
  fill(begin(dp), end(dp), 0);
  dp[0] = 1;
  for (long long x : v) {
    for (long long num = (long long)n; num >= (long long)1; num--) {
      dp[num] = (dp[num] + dp[num - 1] * x) % mod;
    }
  }
  return dp[k];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tc;
  cin >> tc;
  while (tc--) {
    cin >> n >> k;
    ans = 0;
    fill(g + 1, g + n + 1, vector<long long>());
    for (long long i = (long long)1; i <= (long long)n - 1; i++) {
      long long u, v;
      cin >> u >> v;
      g[u].emplace_back(v);
      g[v].emplace_back(u);
    }
    if (k == 1) {
      cout << n << '\n';
      continue;
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << '\n';
      continue;
    }
    for (long long i = (long long)1; i <= (long long)n; i++) {
      for (long long d = (long long)0; d <= (long long)n; d++) dis[d].clear();
      for (long long j : g[i]) {
        fill(begin(cnt), end(cnt), 0);
        dfs(j, i, 1);
        for (long long d = (long long)1; d <= (long long)n; d++) {
          dis[d].emplace_back(cnt[d]);
        }
      }
      for (long long d = (long long)1; d <= (long long)n; d++) {
        ans += get(dis[d]);
        ans %= mod;
      }
    }
    cout << ans << '\n';
  }
}

```