#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e2 + 5;
const long long mod = 1e9 + 7;
const long long INF = 1e9 + 9;
vector<long long> g[maxn];
long long dp[maxn][maxn];
long long depth[maxn][maxn];
long long maxdep, cc;
void dfs(long long x, long long p, long long t, long long level) {
  maxdep = max(maxdep, level);
  if (t == -1) {
    long long tot = 0;
    for (auto i : g[x]) dfs(i, x, ++tot, 1);
    cc = tot;
  } else {
    depth[t][level]++;
    for (auto i : g[x]) {
      if (i == p) continue;
      dfs(i, x, t, level + 1);
    }
  }
}
long long rundp(long long n, long long level, long long k) {
  dp[0][0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long c = 0; c <= cc; c++) {
      if (c == 0)
        dp[i][c] = dp[i - 1][c];
      else
        dp[i][c] =
            (dp[i - 1][c] + dp[i - 1][c - 1] * depth[i][level] % mod) % mod;
    }
  }
  return dp[cc][k];
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k, ans = 0;
    cin >> n >> k;
    for (long long i = 1; i <= n; i++) g[i].clear();
    for (long long i = 1; i < n; i++) {
      long long u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << '\n';
      continue;
    }
    for (long long i = 1; i <= n; i++) {
      maxdep = 0;
      memset(depth, 0, sizeof(depth));
      dfs(i, -1, -1, 0);
      for (long long j = 1; j <= maxdep; j++) {
        memset(dp, 0, sizeof(dp));
        ans = (rundp(n, j, k) + ans) % mod;
      }
    }
    cout << ans << '\n';
  }
}

### 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;
const long long maxn = 1e2 + 5;
const long long mod = 1e9 + 7;
const long long INF = 1e9 + 9;
vector<long long> g[maxn];
long long dp[maxn][maxn];
long long depth[maxn][maxn];
long long maxdep, cc;
void dfs(long long x, long long p, long long t, long long level) {
  maxdep = max(maxdep, level);
  if (t == -1) {
    long long tot = 0;
    for (auto i : g[x]) dfs(i, x, ++tot, 1);
    cc = tot;
  } else {
    depth[t][level]++;
    for (auto i : g[x]) {
      if (i == p) continue;
      dfs(i, x, t, level + 1);
    }
  }
}
long long rundp(long long n, long long level, long long k) {
  dp[0][0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long c = 0; c <= cc; c++) {
      if (c == 0)
        dp[i][c] = dp[i - 1][c];
      else
        dp[i][c] =
            (dp[i - 1][c] + dp[i - 1][c - 1] * depth[i][level] % mod) % mod;
    }
  }
  return dp[cc][k];
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k, ans = 0;
    cin >> n >> k;
    for (long long i = 1; i <= n; i++) g[i].clear();
    for (long long i = 1; i < n; i++) {
      long long u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << '\n';
      continue;
    }
    for (long long i = 1; i <= n; i++) {
      maxdep = 0;
      memset(depth, 0, sizeof(depth));
      dfs(i, -1, -1, 0);
      for (long long j = 1; j <= maxdep; j++) {
        memset(dp, 0, sizeof(dp));
        ans = (rundp(n, j, k) + ans) % mod;
      }
    }
    cout << ans << '\n';
  }
}

```