#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const long long N = 1e3;
const long long MOD = 1e9 + 7;
const long long inf = 1e10;
const long double eps = 1e-12;
const long double EPS = 1e-10;
long long rx[] = {1, -1, 0, 0};
long long ry[] = {0, 0, 1, -1};
vector<long long> g[N];
long long used[N], dp[N][N], cnt[N];
long long rundp(long long m, long long k) {
  for (long long i = 0; i <= m; ++i) {
    for (long long j = 0; j <= k; ++j) {
      dp[i][j] = 0;
    }
  }
  dp[0][0] = 1;
  for (long long i = 0; i < m; ++i) {
    for (long long j = 0; j <= k; ++j) {
      dp[i + 1][j] += dp[i][j];
      dp[i + 1][j] %= MOD;
      dp[i + 1][j + 1] += (dp[i][j] * cnt[i]);
      dp[i + 1][j + 1] %= MOD;
    }
  }
  return dp[m][k];
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long Q;
  cin >> Q;
  while (Q--) {
    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;
      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 ans = 0;
    for (long long c = 0; c < n; ++c) {
      fill(used, used + n, 0);
      used[c] = 1;
      vector<pair<long long, long long> > layer;
      long long m = g[c].size();
      for (long long i = 0; i < m; ++i) {
        long long x = g[c][i];
        layer.push_back({x, i});
        cnt[i] = 1;
        used[x] = 1;
      }
      while (!layer.empty()) {
        ans = ans + rundp(m, k);
        ans %= MOD;
        vector<pair<long long, long long> > nlayer;
        for (auto to : layer) {
          cnt[to.second]--;
          for (auto x : g[to.first]) {
            if (used[x]) continue;
            nlayer.push_back({x, to.second});
            used[x] = 1;
            cnt[to.second]++;
          }
        }
        layer = nlayer;
      }
    }
    cout << ans << '\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>
#pragma GCC optimize("Ofast")
using namespace std;
const long long N = 1e3;
const long long MOD = 1e9 + 7;
const long long inf = 1e10;
const long double eps = 1e-12;
const long double EPS = 1e-10;
long long rx[] = {1, -1, 0, 0};
long long ry[] = {0, 0, 1, -1};
vector<long long> g[N];
long long used[N], dp[N][N], cnt[N];
long long rundp(long long m, long long k) {
  for (long long i = 0; i <= m; ++i) {
    for (long long j = 0; j <= k; ++j) {
      dp[i][j] = 0;
    }
  }
  dp[0][0] = 1;
  for (long long i = 0; i < m; ++i) {
    for (long long j = 0; j <= k; ++j) {
      dp[i + 1][j] += dp[i][j];
      dp[i + 1][j] %= MOD;
      dp[i + 1][j + 1] += (dp[i][j] * cnt[i]);
      dp[i + 1][j + 1] %= MOD;
    }
  }
  return dp[m][k];
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long Q;
  cin >> Q;
  while (Q--) {
    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;
      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 ans = 0;
    for (long long c = 0; c < n; ++c) {
      fill(used, used + n, 0);
      used[c] = 1;
      vector<pair<long long, long long> > layer;
      long long m = g[c].size();
      for (long long i = 0; i < m; ++i) {
        long long x = g[c][i];
        layer.push_back({x, i});
        cnt[i] = 1;
        used[x] = 1;
      }
      while (!layer.empty()) {
        ans = ans + rundp(m, k);
        ans %= MOD;
        vector<pair<long long, long long> > nlayer;
        for (auto to : layer) {
          cnt[to.second]--;
          for (auto x : g[to.first]) {
            if (used[x]) continue;
            nlayer.push_back({x, to.second});
            used[x] = 1;
            cnt[to.second]++;
          }
        }
        layer = nlayer;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

```