#include <bits/stdc++.h>
using namespace std;
const long long p = 1e9 + 7;
const long long maxn = 105;
vector<long long> a[maxn];
bool used[maxn];
long long l[maxn][maxn];
long long ans;
long long c;
long long k;
long long slv1(vector<long long> v, long long k) {
  long long dp[v.size() + 1][k + 1];
  for (long long i = 0; i <= v.size(); ++i)
    for (long long j = 0; j <= k; ++j) dp[i][j] = 0;
  dp[0][0] = 1;
  for (long long i = 0; i < v.size(); ++i)
    for (long long j = 0; j <= k; ++j) {
      if (j != k) {
        dp[i + 1][j + 1] += (dp[i][j] * v[i]);
        dp[i + 1][j + 1] %= p;
      }
      {
        dp[i + 1][j] += (dp[i][j]);
        dp[i + 1][j] %= p;
      }
    }
  return dp[v.size()][k];
}
void dfs(long long x) {
  for (long long i = 0; i < maxn; ++i) l[x][i] = 0;
  l[x][0]++;
  used[x] = true;
  vector<long long> z;
  for (auto v : a[x]) {
    if (used[v]) continue;
    dfs(v);
    z.push_back(v);
    for (long long i = 0; i < maxn; ++i) l[x][i + 1] += l[v][i];
  }
  used[x] = false;
  if (x == c) {
    for (long long i = 0; i < maxn; ++i) {
      vector<long long> v;
      for (auto h : z) {
        v.push_back(l[h][i]);
      }
      long long o = slv1(v, k);
      ans += o;
      ans %= p;
      ans += p;
      ans %= p;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n >> k;
    ans = 0;
    for (long long i = 0; i < maxn; ++i) {
      a[i].clear();
      used[i] = false;
    }
    for (long long i = 0; i < (n - 1); ++i) {
      long long x, y;
      cin >> x >> y;
      x--;
      y--;
      a[x].push_back(y);
      a[y].push_back(x);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << endl;
      continue;
    }
    for (long long i = 0; i < n; ++i) {
      c = i;
      dfs(i);
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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 long long p = 1e9 + 7;
const long long maxn = 105;
vector<long long> a[maxn];
bool used[maxn];
long long l[maxn][maxn];
long long ans;
long long c;
long long k;
long long slv1(vector<long long> v, long long k) {
  long long dp[v.size() + 1][k + 1];
  for (long long i = 0; i <= v.size(); ++i)
    for (long long j = 0; j <= k; ++j) dp[i][j] = 0;
  dp[0][0] = 1;
  for (long long i = 0; i < v.size(); ++i)
    for (long long j = 0; j <= k; ++j) {
      if (j != k) {
        dp[i + 1][j + 1] += (dp[i][j] * v[i]);
        dp[i + 1][j + 1] %= p;
      }
      {
        dp[i + 1][j] += (dp[i][j]);
        dp[i + 1][j] %= p;
      }
    }
  return dp[v.size()][k];
}
void dfs(long long x) {
  for (long long i = 0; i < maxn; ++i) l[x][i] = 0;
  l[x][0]++;
  used[x] = true;
  vector<long long> z;
  for (auto v : a[x]) {
    if (used[v]) continue;
    dfs(v);
    z.push_back(v);
    for (long long i = 0; i < maxn; ++i) l[x][i + 1] += l[v][i];
  }
  used[x] = false;
  if (x == c) {
    for (long long i = 0; i < maxn; ++i) {
      vector<long long> v;
      for (auto h : z) {
        v.push_back(l[h][i]);
      }
      long long o = slv1(v, k);
      ans += o;
      ans %= p;
      ans += p;
      ans %= p;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n >> k;
    ans = 0;
    for (long long i = 0; i < maxn; ++i) {
      a[i].clear();
      used[i] = false;
    }
    for (long long i = 0; i < (n - 1); ++i) {
      long long x, y;
      cin >> x >> y;
      x--;
      y--;
      a[x].push_back(y);
      a[y].push_back(x);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << endl;
      continue;
    }
    for (long long i = 0; i < n; ++i) {
      c = i;
      dfs(i);
    }
    cout << ans << endl;
  }
  return 0;
}

```