#include <bits/stdc++.h>
using namespace std;
const int N = 105, mod = 1e9 + 7;
int T, n, m, x, y, soncnt, cnt[N][N], dep[N], son[N];
long long ans;
vector<int> G[N];
long long f[N][N];
inline void dfs(int x, int fa, int col) {
  dep[x] = dep[fa] + 1, ++cnt[col][dep[x]];
  for (auto y : G[x])
    if (y != fa) dfs(y, x, col);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> T;
  while (T--) {
    cin >> n >> m, ans = 0;
    for (int i = 1; i <= n; ++i) G[i].clear();
    for (int i = 1; i < n; ++i) {
      cin >> x >> y;
      G[x].push_back(y);
      G[y].push_back(x);
    }
    if (m == 2) {
      cout << (n * (n - 1) / 2) << endl;
      continue;
    }
    for (int i = 1; i <= n; ++i) {
      memset(cnt, 0, sizeof cnt);
      dep[i] = soncnt = 0;
      for (auto y : G[i]) dfs(y, i, y), son[++soncnt] = y;
      for (int D = 1; D <= n; ++D) {
        memset(f, 0, sizeof f);
        f[0][0] = 1;
        for (int j = 1; j <= soncnt; ++j) {
          f[j][0] = f[j - 1][0];
          for (int k = 1; k <= j; ++k)
            f[j][k] =
                (f[j - 1][k - 1] * cnt[son[j]][D] % mod + f[j - 1][k]) % mod;
        }
        (ans += f[soncnt][m]) %= mod;
      }
    }
    cout << ans << endl;
  }
  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;
const int N = 105, mod = 1e9 + 7;
int T, n, m, x, y, soncnt, cnt[N][N], dep[N], son[N];
long long ans;
vector<int> G[N];
long long f[N][N];
inline void dfs(int x, int fa, int col) {
  dep[x] = dep[fa] + 1, ++cnt[col][dep[x]];
  for (auto y : G[x])
    if (y != fa) dfs(y, x, col);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> T;
  while (T--) {
    cin >> n >> m, ans = 0;
    for (int i = 1; i <= n; ++i) G[i].clear();
    for (int i = 1; i < n; ++i) {
      cin >> x >> y;
      G[x].push_back(y);
      G[y].push_back(x);
    }
    if (m == 2) {
      cout << (n * (n - 1) / 2) << endl;
      continue;
    }
    for (int i = 1; i <= n; ++i) {
      memset(cnt, 0, sizeof cnt);
      dep[i] = soncnt = 0;
      for (auto y : G[i]) dfs(y, i, y), son[++soncnt] = y;
      for (int D = 1; D <= n; ++D) {
        memset(f, 0, sizeof f);
        f[0][0] = 1;
        for (int j = 1; j <= soncnt; ++j) {
          f[j][0] = f[j - 1][0];
          for (int k = 1; k <= j; ++k)
            f[j][k] =
                (f[j - 1][k - 1] * cnt[son[j]][D] % mod + f[j - 1][k]) % mod;
        }
        (ans += f[soncnt][m]) %= mod;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```