#include <bits/stdc++.h>
using namespace std;
const long long nax = 125, mod = 1e9 + 7;
long long add(long long x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
long long mul(long long x, long long y) { return x * y % mod; }
vector<long long> g[nax];
long long vis[nax], cnt[nax], dp[nax][nax];
long long find(long long n, long long k) {
  for (long long i = 0; i <= n; i++)
    for (long long j = 0; j <= k; j++) dp[i][j] = 0;
  dp[0][0] = 1;
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j <= k; j++) {
      dp[i + 1][j] = add(dp[i + 1][j], dp[i][j]);
      dp[i + 1][j + 1] = add(dp[i + 1][j + 1], mul(dp[i][j], cnt[i]));
    }
  return dp[n][k];
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    for (long long i = 0; i < n; i++) g[i].clear();
    for (long long i = 1; i < n; i++) {
      long long x, y;
      cin >> x >> y;
      g[--x].push_back(--y);
      g[y].push_back(x);
    }
    if (k == 2) {
      cout << (n * (n - 1) / 2) % mod << '\n';
      continue;
    }
    long long res = 0;
    for (long long x = 0; x < n; x++) {
      memset(vis, 0, sizeof(vis));
      vis[x] = 1;
      vector<pair<long long, long long> > e;
      long long m = g[x].size();
      for (long long i = 0; i < m; i++) {
        long long y = g[x][i];
        e.push_back({y, i});
        cnt[i] = 1;
        vis[y] = 1;
      }
      while (!e.empty()) {
        res = add(res, find(m, k));
        vector<pair<long long, long long> > ee;
        for (auto p : e) {
          cnt[p.second]--;
          for (auto y : g[p.first])
            if (!vis[y]) {
              ee.push_back({y, p.second});
              vis[y] = 1;
              cnt[p.second]++;
            }
        }
        e = ee;
      }
    }
    cout << res << '\n';
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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 long long nax = 125, mod = 1e9 + 7;
long long add(long long x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
long long mul(long long x, long long y) { return x * y % mod; }
vector<long long> g[nax];
long long vis[nax], cnt[nax], dp[nax][nax];
long long find(long long n, long long k) {
  for (long long i = 0; i <= n; i++)
    for (long long j = 0; j <= k; j++) dp[i][j] = 0;
  dp[0][0] = 1;
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j <= k; j++) {
      dp[i + 1][j] = add(dp[i + 1][j], dp[i][j]);
      dp[i + 1][j + 1] = add(dp[i + 1][j + 1], mul(dp[i][j], cnt[i]));
    }
  return dp[n][k];
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    for (long long i = 0; i < n; i++) g[i].clear();
    for (long long i = 1; i < n; i++) {
      long long x, y;
      cin >> x >> y;
      g[--x].push_back(--y);
      g[y].push_back(x);
    }
    if (k == 2) {
      cout << (n * (n - 1) / 2) % mod << '\n';
      continue;
    }
    long long res = 0;
    for (long long x = 0; x < n; x++) {
      memset(vis, 0, sizeof(vis));
      vis[x] = 1;
      vector<pair<long long, long long> > e;
      long long m = g[x].size();
      for (long long i = 0; i < m; i++) {
        long long y = g[x][i];
        e.push_back({y, i});
        cnt[i] = 1;
        vis[y] = 1;
      }
      while (!e.empty()) {
        res = add(res, find(m, k));
        vector<pair<long long, long long> > ee;
        for (auto p : e) {
          cnt[p.second]--;
          for (auto y : g[p.first])
            if (!vis[y]) {
              ee.push_back({y, p.second});
              vis[y] = 1;
              cnt[p.second]++;
            }
        }
        e = ee;
      }
    }
    cout << res << '\n';
  }
  return 0;
}

```