#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int mxN = 1e3 + 5, mod = 1e9 + 7, LOG = 20;
int n, k, cnt[mxN], ans;
int dp[mxN][mxN];
vector<int> adj[mxN];
void calc(int v, int p, int dep = 1) {
  ++cnt[dep];
  for (auto to : adj[v]) {
    if (to != p) calc(to, v, dep + 1);
  }
}
void get(int node) {
  for (int i = 0; i <= n; ++i) {
    for (int j = 0; j <= n; ++j) dp[i][j] = 0;
  }
  for (int i = 1; i <= n; ++i) dp[0][i] = 1;
  int now = 0;
  for (auto to : adj[node]) {
    calc(to, node);
    ++now;
    for (int d = 1; d <= n; ++d) {
      for (int j = now; j > 0; j--)
        dp[j][d] = (dp[j][d] + dp[j - 1][d] * 1ll * cnt[d] % mod) % mod;
    }
    for (int i = 1; i <= n; ++i) cnt[i] = 0;
  }
  for (int i = 1; i <= n; ++i) ans = (ans + dp[k][i]) % mod;
}
void solve() {
  cin >> n >> k;
  ans = 0;
  for (int i = 1; i <= n; ++i) adj[i].clear();
  for (int i = 1, u, v; i < n; ++i) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  if (k == 2) {
    cout << (n * 1ll * (n - 1) / 2) % mod;
    return;
  }
  for (int c = 1; c <= n; ++c) {
    get(c);
  }
  cout << ans;
}
int main() {
  ios ::sync_with_stdio(false), cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    solve();
    cout << '\n';
  }
  return 0;
}

### Prompt

Generate 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;
using ll = long long;
const int mxN = 1e3 + 5, mod = 1e9 + 7, LOG = 20;
int n, k, cnt[mxN], ans;
int dp[mxN][mxN];
vector<int> adj[mxN];
void calc(int v, int p, int dep = 1) {
  ++cnt[dep];
  for (auto to : adj[v]) {
    if (to != p) calc(to, v, dep + 1);
  }
}
void get(int node) {
  for (int i = 0; i <= n; ++i) {
    for (int j = 0; j <= n; ++j) dp[i][j] = 0;
  }
  for (int i = 1; i <= n; ++i) dp[0][i] = 1;
  int now = 0;
  for (auto to : adj[node]) {
    calc(to, node);
    ++now;
    for (int d = 1; d <= n; ++d) {
      for (int j = now; j > 0; j--)
        dp[j][d] = (dp[j][d] + dp[j - 1][d] * 1ll * cnt[d] % mod) % mod;
    }
    for (int i = 1; i <= n; ++i) cnt[i] = 0;
  }
  for (int i = 1; i <= n; ++i) ans = (ans + dp[k][i]) % mod;
}
void solve() {
  cin >> n >> k;
  ans = 0;
  for (int i = 1; i <= n; ++i) adj[i].clear();
  for (int i = 1, u, v; i < n; ++i) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  if (k == 2) {
    cout << (n * 1ll * (n - 1) / 2) % mod;
    return;
  }
  for (int c = 1; c <= n; ++c) {
    get(c);
  }
  cout << ans;
}
int main() {
  ios ::sync_with_stdio(false), cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    solve();
    cout << '\n';
  }
  return 0;
}

```