#include <bits/stdc++.h>
using namespace std;
const int N = 105;
int n, d[N], k, vis[N][N], dp[N][N], id, mod = 1e9 + 7, cnt[N][N];
vector<int> adj[N], v;
void dfs(int node, int par, int parCh) {
  if (parCh) {
    cnt[parCh][d[node]]++;
  }
  for (auto ch : adj[node]) {
    if (ch == par) continue;
    d[ch] = d[node] + 1;
    if (node == par)
      dfs(ch, node, ch);
    else
      dfs(ch, node, parCh);
  }
}
int solve(int i, int rem) {
  if (!rem) return 1;
  if (i == (int)v.size()) return 0;
  if (vis[i][rem] == id) return dp[i][rem];
  vis[i][rem] = id;
  int ans = solve(i + 1, rem);
  if (v[i]) ans = (ans + 1ll * v[i] * solve(i + 1, rem - 1)) % mod;
  return dp[i][rem] = ans;
}
void clear() {
  for (int i = 0; i <= n; i++) {
    adj[i].clear();
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    for (int i = 1; i < n; i++) {
      int x, y;
      cin >> x >> y;
      adj[x].push_back(y);
      adj[y].push_back(x);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << '\n';
      clear();
      continue;
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      memset(d, 0, sizeof(d));
      memset(cnt, 0, sizeof(cnt));
      dfs(i, i, 0);
      for (int ds = 1; ds <= n; ds++) {
        v.clear();
        for (int ch = 1; ch <= n; ch++) {
          v.push_back(cnt[ch][ds]);
        }
        id++;
        ans = (ans + solve(0, k)) % mod;
      }
    }
    cout << ans << '\n';
    clear();
  }
  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 N = 105;
int n, d[N], k, vis[N][N], dp[N][N], id, mod = 1e9 + 7, cnt[N][N];
vector<int> adj[N], v;
void dfs(int node, int par, int parCh) {
  if (parCh) {
    cnt[parCh][d[node]]++;
  }
  for (auto ch : adj[node]) {
    if (ch == par) continue;
    d[ch] = d[node] + 1;
    if (node == par)
      dfs(ch, node, ch);
    else
      dfs(ch, node, parCh);
  }
}
int solve(int i, int rem) {
  if (!rem) return 1;
  if (i == (int)v.size()) return 0;
  if (vis[i][rem] == id) return dp[i][rem];
  vis[i][rem] = id;
  int ans = solve(i + 1, rem);
  if (v[i]) ans = (ans + 1ll * v[i] * solve(i + 1, rem - 1)) % mod;
  return dp[i][rem] = ans;
}
void clear() {
  for (int i = 0; i <= n; i++) {
    adj[i].clear();
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    for (int i = 1; i < n; i++) {
      int x, y;
      cin >> x >> y;
      adj[x].push_back(y);
      adj[y].push_back(x);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << '\n';
      clear();
      continue;
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      memset(d, 0, sizeof(d));
      memset(cnt, 0, sizeof(cnt));
      dfs(i, i, 0);
      for (int ds = 1; ds <= n; ds++) {
        v.clear();
        for (int ch = 1; ch <= n; ch++) {
          v.push_back(cnt[ch][ds]);
        }
        id++;
        ans = (ans + solve(0, k)) % mod;
      }
    }
    cout << ans << '\n';
    clear();
  }
  return 0;
}

```