#include <bits/stdc++.h>
using namespace std;
using i64 = int64_t;
using _ii = pair<int, int>;
const int N = 107;
int t, n, k;
i64 dp_in[N][N], dp_out[N][N], ans, mod = 1e9 + 7;
vector<vector<int>> a;
void DFS(int h, int legacy) {
  for (auto &i : a[h]) {
    if (i == legacy) continue;
    DFS(i, h);
    for (int j = 1; j <= n; j++)
      dp_in[h][j] = (dp_in[h][j] + dp_in[i][j - 1]) % mod;
  }
  ++dp_in[h][0];
}
void Cal(vector<int> &a, int cnt) {
  int m = a.size();
  vector<i64> f(cnt + 1);
  f[0] = 1;
  for (auto &i : a) {
    for (int j = cnt; j >= 1; j--) {
      f[j] += f[j - 1] * i;
      f[j] %= mod;
    }
  }
  ans = (ans + f[cnt]) % mod;
}
void Run(int h, int legacy) {
  for (int j = 1; j <= n; j++) {
    vector<int> d;
    d.push_back(dp_out[h][j]);
    for (auto &i : a[h]) {
      if (i == legacy) continue;
      d.push_back(dp_in[i][j - 1]);
    }
    Cal(d, k);
  }
  for (auto &i : a[h]) {
    if (i == legacy) continue;
    for (int j = 1; j <= n; j++) {
      int temp = dp_in[h][j - 1] - (j - 2 >= 0 ? dp_in[i][j - 2] : 0);
      dp_out[i][j] = temp + dp_out[h][j - 1];
    }
    Run(i, h);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> k;
    a.resize(n + 1);
    for (auto &i : a) i.clear();
    for (int i = 1; i <= n - 1; i++) {
      int x, y;
      cin >> x >> y;
      a[x].push_back(y);
      a[y].push_back(x);
    }
    ans = 0;
    memset(dp_in, 0, sizeof(dp_in));
    memset(dp_out, 0, sizeof(dp_out));
    DFS(1, 0);
    Run(1, 0);
    if (k == 2) ans = (n * (n - 1) / 2) % 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;
using i64 = int64_t;
using _ii = pair<int, int>;
const int N = 107;
int t, n, k;
i64 dp_in[N][N], dp_out[N][N], ans, mod = 1e9 + 7;
vector<vector<int>> a;
void DFS(int h, int legacy) {
  for (auto &i : a[h]) {
    if (i == legacy) continue;
    DFS(i, h);
    for (int j = 1; j <= n; j++)
      dp_in[h][j] = (dp_in[h][j] + dp_in[i][j - 1]) % mod;
  }
  ++dp_in[h][0];
}
void Cal(vector<int> &a, int cnt) {
  int m = a.size();
  vector<i64> f(cnt + 1);
  f[0] = 1;
  for (auto &i : a) {
    for (int j = cnt; j >= 1; j--) {
      f[j] += f[j - 1] * i;
      f[j] %= mod;
    }
  }
  ans = (ans + f[cnt]) % mod;
}
void Run(int h, int legacy) {
  for (int j = 1; j <= n; j++) {
    vector<int> d;
    d.push_back(dp_out[h][j]);
    for (auto &i : a[h]) {
      if (i == legacy) continue;
      d.push_back(dp_in[i][j - 1]);
    }
    Cal(d, k);
  }
  for (auto &i : a[h]) {
    if (i == legacy) continue;
    for (int j = 1; j <= n; j++) {
      int temp = dp_in[h][j - 1] - (j - 2 >= 0 ? dp_in[i][j - 2] : 0);
      dp_out[i][j] = temp + dp_out[h][j - 1];
    }
    Run(i, h);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> k;
    a.resize(n + 1);
    for (auto &i : a) i.clear();
    for (int i = 1; i <= n - 1; i++) {
      int x, y;
      cin >> x >> y;
      a[x].push_back(y);
      a[y].push_back(x);
    }
    ans = 0;
    memset(dp_in, 0, sizeof(dp_in));
    memset(dp_out, 0, sizeof(dp_out));
    DFS(1, 0);
    Run(1, 0);
    if (k == 2) ans = (n * (n - 1) / 2) % mod;
    cout << ans << '\n';
  }
}

```