#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long N = 1e2 + 10;
vector<long long> g[N];
void dfs(long long node, long long par, long long dd, vector<long long>& fre) {
  fre[dd]++;
  for (auto x : g[node])
    if (x != par) dfs(x, node, dd + 1, fre);
}
long long n, k;
long long get(vector<long long>& arr) {
  if (arr.size() < k) return 0;
  vector<long long> cur(k + 1, 0);
  cur[0] = 1;
  for (auto x : arr) {
    vector<long long> nex(k + 1, 0);
    for (long long i = 0; i <= k; i++) {
      nex[i] = cur[i];
      if (i > 0) nex[i] += cur[i - 1] * x;
      nex[i] %= MOD;
    }
    swap(cur, nex);
  }
  return cur[k];
}
long long solve(long long x) {
  long long sz = g[x].size();
  vector<vector<long long> > go(sz, vector<long long>(n + 1, 0));
  for (long long i = 0; i < sz; i++) dfs(g[x][i], x, 1, go[i]);
  long long res = 0;
  for (long long i = 1; i <= n; i++) {
    vector<long long> arr;
    for (auto v : go)
      if (v[i] > 0) arr.push_back(v[i]);
    if (arr.empty()) break;
    res += get(arr);
    res %= MOD;
  }
  return res;
}
void solve() {
  cin >> n >> k;
  long long u, v;
  for (long long i = 0; i < n - 1; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  long long res = 0;
  if (k == 2) {
    res = (n * (n - 1)) / 2;
    res %= MOD;
  } else {
    for (long long i = 1; i <= n; i++) {
      res += solve(i);
      res %= MOD;
    }
  }
  cout << res << '\n';
  for (long long i = 1; i <= n; i++) g[i].clear();
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) solve();
}

### 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>
using namespace std;
const long long MOD = 1e9 + 7;
const long long N = 1e2 + 10;
vector<long long> g[N];
void dfs(long long node, long long par, long long dd, vector<long long>& fre) {
  fre[dd]++;
  for (auto x : g[node])
    if (x != par) dfs(x, node, dd + 1, fre);
}
long long n, k;
long long get(vector<long long>& arr) {
  if (arr.size() < k) return 0;
  vector<long long> cur(k + 1, 0);
  cur[0] = 1;
  for (auto x : arr) {
    vector<long long> nex(k + 1, 0);
    for (long long i = 0; i <= k; i++) {
      nex[i] = cur[i];
      if (i > 0) nex[i] += cur[i - 1] * x;
      nex[i] %= MOD;
    }
    swap(cur, nex);
  }
  return cur[k];
}
long long solve(long long x) {
  long long sz = g[x].size();
  vector<vector<long long> > go(sz, vector<long long>(n + 1, 0));
  for (long long i = 0; i < sz; i++) dfs(g[x][i], x, 1, go[i]);
  long long res = 0;
  for (long long i = 1; i <= n; i++) {
    vector<long long> arr;
    for (auto v : go)
      if (v[i] > 0) arr.push_back(v[i]);
    if (arr.empty()) break;
    res += get(arr);
    res %= MOD;
  }
  return res;
}
void solve() {
  cin >> n >> k;
  long long u, v;
  for (long long i = 0; i < n - 1; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  long long res = 0;
  if (k == 2) {
    res = (n * (n - 1)) / 2;
    res %= MOD;
  } else {
    for (long long i = 1; i <= n; i++) {
      res += solve(i);
      res %= MOD;
    }
  }
  cout << res << '\n';
  for (long long i = 1; i <= n; i++) g[i].clear();
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) solve();
}

```