#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,tune=native")
template <typename T>
using Prior = std::priority_queue<T>;
template <typename T>
using prior = std::priority_queue<T, std::vector<T>, std::greater<T>>;
using namespace std;
const long long md = 1e9 + 7;
inline long long mod(long long x) { return (x % md + md) % md; }
int n, k;
vector<vector<long long>> g;
int tra(int src, int dir, int step) {
  if (!step) return 1;
  int ret = 0;
  for (auto nbr : g[dir]) {
    if (nbr == src) continue;
    ret += tra(dir, nbr, step - 1);
  }
  return ret;
}
long long h(const vector<long long>& cef, int k) {
  const int m = cef.size();
  assert(m >= k);
  vector<vector<long long>> dp(m + 1, vector<long long>(k + 1));
  for (int i = 0; i <= m; ++i) dp[i][0] = 1;
  for (int i = 1; i <= m; ++i)
    for (int j = 1; j <= min(i, k); ++j)
      dp[i][j] = mod(dp[i - 1][j] + mod(dp[i - 1][j - 1] * cef[i - 1]));
  return dp[m][k];
}
void solve() {
  cin >> n >> k;
  g = vector<vector<long long>>(n);
  for (auto [u, v] : vector<pair<long long, long long>>(n - 1)) {
    cin >> u >> v, --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  };
  ;
  if (k == 2) return cout << (n * (n - 1LL) >> 1) << '\n', void();
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    int m = g[i].size();
    for (int step = 1; step < n; ++step) {
      vector<long long> cnt;
      for (auto dir : g[i])
        if (int tmp = tra(i, dir, step - 1)) cnt.push_back(tmp);
      if (cnt.size() < k) continue;
      ans = mod(ans + h(cnt, k));
    }
  }
  return cout << ans << '\n', void();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int _ = 1;
  cin >> _;
  vector<long long> __(_);
  iota(begin(__), end(__), 1);
  for (auto i : __) {
    solve();
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,tune=native")
template <typename T>
using Prior = std::priority_queue<T>;
template <typename T>
using prior = std::priority_queue<T, std::vector<T>, std::greater<T>>;
using namespace std;
const long long md = 1e9 + 7;
inline long long mod(long long x) { return (x % md + md) % md; }
int n, k;
vector<vector<long long>> g;
int tra(int src, int dir, int step) {
  if (!step) return 1;
  int ret = 0;
  for (auto nbr : g[dir]) {
    if (nbr == src) continue;
    ret += tra(dir, nbr, step - 1);
  }
  return ret;
}
long long h(const vector<long long>& cef, int k) {
  const int m = cef.size();
  assert(m >= k);
  vector<vector<long long>> dp(m + 1, vector<long long>(k + 1));
  for (int i = 0; i <= m; ++i) dp[i][0] = 1;
  for (int i = 1; i <= m; ++i)
    for (int j = 1; j <= min(i, k); ++j)
      dp[i][j] = mod(dp[i - 1][j] + mod(dp[i - 1][j - 1] * cef[i - 1]));
  return dp[m][k];
}
void solve() {
  cin >> n >> k;
  g = vector<vector<long long>>(n);
  for (auto [u, v] : vector<pair<long long, long long>>(n - 1)) {
    cin >> u >> v, --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  };
  ;
  if (k == 2) return cout << (n * (n - 1LL) >> 1) << '\n', void();
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    int m = g[i].size();
    for (int step = 1; step < n; ++step) {
      vector<long long> cnt;
      for (auto dir : g[i])
        if (int tmp = tra(i, dir, step - 1)) cnt.push_back(tmp);
      if (cnt.size() < k) continue;
      ans = mod(ans + h(cnt, k));
    }
  }
  return cout << ans << '\n', void();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int _ = 1;
  cin >> _;
  vector<long long> __(_);
  iota(begin(__), end(__), 1);
  for (auto i : __) {
    solve();
  }
  return 0;
}

```