#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 7;
long long mod = 1e9 + 7;
int n, k;
vector<int> e[maxn];
int f[102][102], num[102][102];
void dfs(int u, int p, int len, int fa) {
  num[fa][len]++;
  for (int v : e[u]) {
    if (v == p) continue;
    dfs(v, u, len + 1, fa);
  }
}
int cal(vector<int> v) {
  memset(f, 0, sizeof(f));
  int res = 0;
  int m = v.size();
  f[0][0] = 1;
  for (int i = 1; i <= m; i++) {
    for (int j = 0; j <= i; j++) {
      if (j)
        f[i][j] = (f[i - 1][j] + 1ll * f[i - 1][j - 1] * v[i - 1]) % mod;
      else
        f[i][j] = f[i - 1][j];
    }
  }
  res = f[m][k] % mod;
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int _;
  cin >> _;
  while (_--) {
    cin >> n >> k;
    for (int i = (1); i <= (n - 1); ++i) {
      int u, v;
      cin >> u >> v;
      e[u].push_back(v), e[v].push_back(u);
    }
    if (k == 2) {
      cout << (n * (n - 1) / 2ll) % mod << endl;
      for (int i = (1); i <= (n); ++i) e[i].clear();
      continue;
    }
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      memset(num, 0, sizeof(num));
      for (int v : e[i]) dfs(v, i, 1, v);
      for (int d = 1; d <= n; d++) {
        vector<int> vv;
        for (int v : e[i]) vv.push_back(num[v][d]);
        ans = (ans + cal(vv)) % mod;
      }
    }
    cout << ans << endl;
    for (int i = (1); i <= (n); ++i) e[i].clear();
  }
  return 0;
}

### 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 int maxn = 2e5 + 7;
long long mod = 1e9 + 7;
int n, k;
vector<int> e[maxn];
int f[102][102], num[102][102];
void dfs(int u, int p, int len, int fa) {
  num[fa][len]++;
  for (int v : e[u]) {
    if (v == p) continue;
    dfs(v, u, len + 1, fa);
  }
}
int cal(vector<int> v) {
  memset(f, 0, sizeof(f));
  int res = 0;
  int m = v.size();
  f[0][0] = 1;
  for (int i = 1; i <= m; i++) {
    for (int j = 0; j <= i; j++) {
      if (j)
        f[i][j] = (f[i - 1][j] + 1ll * f[i - 1][j - 1] * v[i - 1]) % mod;
      else
        f[i][j] = f[i - 1][j];
    }
  }
  res = f[m][k] % mod;
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int _;
  cin >> _;
  while (_--) {
    cin >> n >> k;
    for (int i = (1); i <= (n - 1); ++i) {
      int u, v;
      cin >> u >> v;
      e[u].push_back(v), e[v].push_back(u);
    }
    if (k == 2) {
      cout << (n * (n - 1) / 2ll) % mod << endl;
      for (int i = (1); i <= (n); ++i) e[i].clear();
      continue;
    }
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      memset(num, 0, sizeof(num));
      for (int v : e[i]) dfs(v, i, 1, v);
      for (int d = 1; d <= n; d++) {
        vector<int> vv;
        for (int v : e[i]) vv.push_back(num[v][d]);
        ans = (ans + cal(vv)) % mod;
      }
    }
    cout << ans << endl;
    for (int i = (1); i <= (n); ++i) e[i].clear();
  }
  return 0;
}

```