#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 7, mod = 1e9 + 7;
long long dp[105][105], ans;
vector<int> G[105], d;
int n, k, mx;
void dfs(int x, int f, int dep) {
  ++d[dep];
  if (dep > mx) mx = dep;
  for (int i : G[x]) {
    if (i == f) continue;
    dfs(i, x, dep + 1);
  }
}
inline void solve() {
  cin >> n >> k;
  ans = 0;
  for (int i = 1; i <= n; ++i) G[i].clear();
  for (int i = 2, x, y; i <= n; ++i) {
    cin >> x >> y;
    G[x].emplace_back(y);
    G[y].emplace_back(x);
  }
  if (k == 2) {
    cout << n * (n - 1) / 2 << '\n';
    return;
  }
  vector<int> t(n + 1);
  for (int root = 1; root <= n; ++root) {
    vector<vector<int> > v;
    mx = 0;
    for (int x : G[root]) {
      d = t;
      dfs(x, root, 1);
      v.emplace_back(d);
    }
    int m = v.size();
    if (m < k) continue;
    for (int i = 1; i <= mx; ++i) {
      vector<int> p;
      int q = 0;
      for (auto &x : v) p.emplace_back(x[i]), q |= x[i];
      if (!q) break;
      for (int j = 1; j <= m; ++j)
        for (int x = 0; x <= k; ++x) dp[j][x] = 0;
      dp[0][0] = 1;
      for (int j = 1; j <= m; ++j) {
        for (int x = 0; x <= k; ++x) dp[j][x] = dp[j - 1][x];
        for (int x = 1; x <= k; ++x)
          dp[j][x] = (dp[j][x] + dp[j - 1][x - 1] * p[j - 1]) % mod;
      }
      ans = (ans + dp[m][k]) % mod;
    }
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int _ = 1;
  cin >> _;
  while (_--) solve();
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
const int maxn = 2e5 + 7, mod = 1e9 + 7;
long long dp[105][105], ans;
vector<int> G[105], d;
int n, k, mx;
void dfs(int x, int f, int dep) {
  ++d[dep];
  if (dep > mx) mx = dep;
  for (int i : G[x]) {
    if (i == f) continue;
    dfs(i, x, dep + 1);
  }
}
inline void solve() {
  cin >> n >> k;
  ans = 0;
  for (int i = 1; i <= n; ++i) G[i].clear();
  for (int i = 2, x, y; i <= n; ++i) {
    cin >> x >> y;
    G[x].emplace_back(y);
    G[y].emplace_back(x);
  }
  if (k == 2) {
    cout << n * (n - 1) / 2 << '\n';
    return;
  }
  vector<int> t(n + 1);
  for (int root = 1; root <= n; ++root) {
    vector<vector<int> > v;
    mx = 0;
    for (int x : G[root]) {
      d = t;
      dfs(x, root, 1);
      v.emplace_back(d);
    }
    int m = v.size();
    if (m < k) continue;
    for (int i = 1; i <= mx; ++i) {
      vector<int> p;
      int q = 0;
      for (auto &x : v) p.emplace_back(x[i]), q |= x[i];
      if (!q) break;
      for (int j = 1; j <= m; ++j)
        for (int x = 0; x <= k; ++x) dp[j][x] = 0;
      dp[0][0] = 1;
      for (int j = 1; j <= m; ++j) {
        for (int x = 0; x <= k; ++x) dp[j][x] = dp[j - 1][x];
        for (int x = 1; x <= k; ++x)
          dp[j][x] = (dp[j][x] + dp[j - 1][x - 1] * p[j - 1]) % mod;
      }
      ans = (ans + dp[m][k]) % mod;
    }
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int _ = 1;
  cin >> _;
  while (_--) solve();
  return 0;
}

```