#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long Fact_Length = 1e5 + 1;
long long Power(long long base, long long expo) {
  long long $result = 1;
  base %= mod;
  while (expo) {
    if (expo % 2 == 1) $result = ($result * base) % mod;
    base = (base * base) % mod;
    expo /= 2;
  }
  return $result;
}
long long Mod_Inv(long long $a) { return Power($a, mod - 2); }
long long Factorial[Fact_Length];
long long Make_Factorial() {
  Factorial[0] = 1;
  for (long long i = 1; i < Fact_Length; i++) {
    Factorial[i] = (i * Factorial[i - 1]) % mod;
  }
  return 0;
}
long long Implement_Make_Factorial = Make_Factorial();
long long nCr(long long $n, long long $r) {
  if ($n < $r || $n < 0 || $r < 0) return 0;
  long long $ans = (Factorial[$n] * Mod_Inv(Factorial[$r])) % mod;
  $ans = ($ans * Mod_Inv(Factorial[$n - $r])) % mod;
  return $ans;
}
const long long N = 101;
vector<long long> adj[N];
long long n, k;
long long dep[N];
void DFS(long long chi, long long par, long long cur) {
  dep[cur]++;
  for (auto x : adj[chi]) {
    if (x == par) continue;
    DFS(x, chi, cur + 1);
  }
}
long long calc(long long b[], long long m) {
  if (m < k) return 0;
  long long dp[m + 1][k + 1];
  memset(dp, 0, sizeof(dp));
  for (long long i = 0; i <= m; i++) dp[i][0] = 1;
  for (long long i = 1; i <= m; i++) {
    for (long long j = 1; j <= k; j++) {
      dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1] * b[i];
      dp[i][j] %= mod;
    }
  }
  return dp[m][k];
}
long long val(long long root) {
  vector<long long> a[n + 1];
  long long m = adj[root].size();
  for (auto x : adj[root]) {
    for (long long i = 1; i <= n; i++) dep[i] = 0;
    DFS(x, root, 1);
    for (long long i = 1; i <= n; i++) a[i].push_back(dep[i]);
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    sort(a[i].begin(), a[i].end());
    long long b[m + 1];
    for (long long j = 1; j <= m; j++) b[j] = a[i][j - 1];
    ans += calc(b, m);
    ans %= mod;
  }
  return ans;
}
void solve() {
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) adj[i].clear();
  for (long long i = 1; i <= n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  if (k == 2) {
    long long ans = (n * (n - 1)) / 2;
    cout << ans << endl;
    return;
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    ans += val(i);
    ans %= mod;
  }
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  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 mod = 1e9 + 7;
const long long Fact_Length = 1e5 + 1;
long long Power(long long base, long long expo) {
  long long $result = 1;
  base %= mod;
  while (expo) {
    if (expo % 2 == 1) $result = ($result * base) % mod;
    base = (base * base) % mod;
    expo /= 2;
  }
  return $result;
}
long long Mod_Inv(long long $a) { return Power($a, mod - 2); }
long long Factorial[Fact_Length];
long long Make_Factorial() {
  Factorial[0] = 1;
  for (long long i = 1; i < Fact_Length; i++) {
    Factorial[i] = (i * Factorial[i - 1]) % mod;
  }
  return 0;
}
long long Implement_Make_Factorial = Make_Factorial();
long long nCr(long long $n, long long $r) {
  if ($n < $r || $n < 0 || $r < 0) return 0;
  long long $ans = (Factorial[$n] * Mod_Inv(Factorial[$r])) % mod;
  $ans = ($ans * Mod_Inv(Factorial[$n - $r])) % mod;
  return $ans;
}
const long long N = 101;
vector<long long> adj[N];
long long n, k;
long long dep[N];
void DFS(long long chi, long long par, long long cur) {
  dep[cur]++;
  for (auto x : adj[chi]) {
    if (x == par) continue;
    DFS(x, chi, cur + 1);
  }
}
long long calc(long long b[], long long m) {
  if (m < k) return 0;
  long long dp[m + 1][k + 1];
  memset(dp, 0, sizeof(dp));
  for (long long i = 0; i <= m; i++) dp[i][0] = 1;
  for (long long i = 1; i <= m; i++) {
    for (long long j = 1; j <= k; j++) {
      dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1] * b[i];
      dp[i][j] %= mod;
    }
  }
  return dp[m][k];
}
long long val(long long root) {
  vector<long long> a[n + 1];
  long long m = adj[root].size();
  for (auto x : adj[root]) {
    for (long long i = 1; i <= n; i++) dep[i] = 0;
    DFS(x, root, 1);
    for (long long i = 1; i <= n; i++) a[i].push_back(dep[i]);
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    sort(a[i].begin(), a[i].end());
    long long b[m + 1];
    for (long long j = 1; j <= m; j++) b[j] = a[i][j - 1];
    ans += calc(b, m);
    ans %= mod;
  }
  return ans;
}
void solve() {
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) adj[i].clear();
  for (long long i = 1; i <= n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  if (k == 2) {
    long long ans = (n * (n - 1)) / 2;
    cout << ans << endl;
    return;
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    ans += val(i);
    ans %= mod;
  }
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```