#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 7;
const int N = 1e2 + 10;
vector<int> v[N];
void dfs(int x, vector<int>& cnt, int fa = 0, int d = 0) {
  cnt[d]++;
  for (auto& y : v[x]) {
    if (y == fa) continue;
    dfs(y, cnt, x, d + 1);
  }
}
vector<int> multy(vector<int>& v1, vector<int>& v2) {
  while (v1.size() > 1 && v1.back() == 0) v1.pop_back();
  while (v2.size() > 1 && v2.back() == 0) v2.pop_back();
  vector<int> ret(v1.size() + v2.size() - 1);
  for (int i = 0; i < v1.size(); i++) {
    for (int j = 0; j < v2.size(); j++) {
      ret[i + j] = (ret[i + j] + 1LL * v1[i] * v2[j]) % INF;
    }
  }
  return ret;
}
int solve(vector<int>& vp, int k) {
  if (vp.size() < k) return 0;
  vector<int> ret;
  ret.push_back(1);
  for (auto& x : vp) {
    vector<int> v{1, x};
    ret = multy(ret, v);
  }
  if (ret.size() <= k)
    return 0;
  else
    return ret[k];
}
int main() {
  int ncase;
  scanf("%d", &ncase);
  while (ncase--) {
    int n, k;
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) v[i].clear();
    for (int i = 1; i < n; i++) {
      int x, y;
      scanf("%d%d", &x, &y);
      v[x].push_back(y);
      v[y].push_back(x);
    }
    if (k == 2) {
      printf("%d\n", n * (n - 1) / 2);
      continue;
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      vector<vector<int>> key(n);
      for (auto& j : v[i]) {
        vector<int> cnt(n, 0);
        dfs(j, cnt, i, 0);
        for (int k = 0; k < n; k++) {
          if (cnt[k] > 0) key[k].push_back(cnt[k]);
        }
      }
      for (auto& vp : key) {
        ans += solve(vp, k);
      }
    }
    printf("%lld\n", ans % INF);
  }
  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>
using namespace std;
const long long INF = 1e9 + 7;
const int N = 1e2 + 10;
vector<int> v[N];
void dfs(int x, vector<int>& cnt, int fa = 0, int d = 0) {
  cnt[d]++;
  for (auto& y : v[x]) {
    if (y == fa) continue;
    dfs(y, cnt, x, d + 1);
  }
}
vector<int> multy(vector<int>& v1, vector<int>& v2) {
  while (v1.size() > 1 && v1.back() == 0) v1.pop_back();
  while (v2.size() > 1 && v2.back() == 0) v2.pop_back();
  vector<int> ret(v1.size() + v2.size() - 1);
  for (int i = 0; i < v1.size(); i++) {
    for (int j = 0; j < v2.size(); j++) {
      ret[i + j] = (ret[i + j] + 1LL * v1[i] * v2[j]) % INF;
    }
  }
  return ret;
}
int solve(vector<int>& vp, int k) {
  if (vp.size() < k) return 0;
  vector<int> ret;
  ret.push_back(1);
  for (auto& x : vp) {
    vector<int> v{1, x};
    ret = multy(ret, v);
  }
  if (ret.size() <= k)
    return 0;
  else
    return ret[k];
}
int main() {
  int ncase;
  scanf("%d", &ncase);
  while (ncase--) {
    int n, k;
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) v[i].clear();
    for (int i = 1; i < n; i++) {
      int x, y;
      scanf("%d%d", &x, &y);
      v[x].push_back(y);
      v[y].push_back(x);
    }
    if (k == 2) {
      printf("%d\n", n * (n - 1) / 2);
      continue;
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      vector<vector<int>> key(n);
      for (auto& j : v[i]) {
        vector<int> cnt(n, 0);
        dfs(j, cnt, i, 0);
        for (int k = 0; k < n; k++) {
          if (cnt[k] > 0) key[k].push_back(cnt[k]);
        }
      }
      for (auto& vp : key) {
        ans += solve(vp, k);
      }
    }
    printf("%lld\n", ans % INF);
  }
  return 0;
}

```