#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const int N = 128;
int t, n, k;
int cnt[N];
bool vis[N];
long long dp[N][N];
vector<int> path[N];
long long calc(int, int);
int main() {
  int a, b;
  scanf("%d", &t);
  while (t--) {
    scanf("%d %d", &n, &k);
    for (int i = 0; i <= n; i++) path[i].clear();
    for (int i = 1; i < n; i++) {
      scanf("%d %d", &a, &b);
      path[a].push_back(b);
      path[b].push_back(a);
    }
    if (k == 2) {
      printf("%lld\n", (n - 1ll) * n / 2 % MOD);
      continue;
    }
    long long ans = 0;
    for (int root = 1; root <= n; root++) {
      memset(vis, false, sizeof(vis));
      vis[root] = true;
      int size = path[root].size();
      vector<pair<int, int>> layer;
      for (int i = 0; i < size; i++) {
        int son = path[root][i];
        layer.emplace_back(son, i);
        cnt[i] = 1;
        vis[son] = true;
      }
      while (!layer.empty()) {
        ans = (ans + calc(size, k)) % MOD;
        vector<pair<int, int>> newLayer;
        for (auto node : layer) {
          cnt[node.second]--;
          for (auto son : path[node.first]) {
            if (vis[son]) continue;
            vis[son] = true;
            cnt[node.second]++;
            newLayer.emplace_back(son, node.second);
          }
        }
        layer = newLayer;
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}
long long calc(int x, int y) {
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 1;
  for (int i = 0; i < x; i++)
    for (int j = 0; j <= y; j++) {
      dp[i + 1][j] = (dp[i + 1][j] + dp[i][j]) % MOD;
      dp[i + 1][j + 1] = (dp[i + 1][j + 1] + dp[i][j] * cnt[i] % MOD) % MOD;
    }
  return dp[x][y];
}

### 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 MOD = 1e9 + 7;
const int N = 128;
int t, n, k;
int cnt[N];
bool vis[N];
long long dp[N][N];
vector<int> path[N];
long long calc(int, int);
int main() {
  int a, b;
  scanf("%d", &t);
  while (t--) {
    scanf("%d %d", &n, &k);
    for (int i = 0; i <= n; i++) path[i].clear();
    for (int i = 1; i < n; i++) {
      scanf("%d %d", &a, &b);
      path[a].push_back(b);
      path[b].push_back(a);
    }
    if (k == 2) {
      printf("%lld\n", (n - 1ll) * n / 2 % MOD);
      continue;
    }
    long long ans = 0;
    for (int root = 1; root <= n; root++) {
      memset(vis, false, sizeof(vis));
      vis[root] = true;
      int size = path[root].size();
      vector<pair<int, int>> layer;
      for (int i = 0; i < size; i++) {
        int son = path[root][i];
        layer.emplace_back(son, i);
        cnt[i] = 1;
        vis[son] = true;
      }
      while (!layer.empty()) {
        ans = (ans + calc(size, k)) % MOD;
        vector<pair<int, int>> newLayer;
        for (auto node : layer) {
          cnt[node.second]--;
          for (auto son : path[node.first]) {
            if (vis[son]) continue;
            vis[son] = true;
            cnt[node.second]++;
            newLayer.emplace_back(son, node.second);
          }
        }
        layer = newLayer;
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}
long long calc(int x, int y) {
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 1;
  for (int i = 0; i < x; i++)
    for (int j = 0; j <= y; j++) {
      dp[i + 1][j] = (dp[i + 1][j] + dp[i][j]) % MOD;
      dp[i + 1][j + 1] = (dp[i + 1][j + 1] + dp[i][j] * cnt[i] % MOD) % MOD;
    }
  return dp[x][y];
}

```