#include <bits/stdc++.h>
using namespace std;
const long long MOD = (1e9 + 7);
const long long MAX_VALEURS = (101);
vector<long long> Voisins[MAX_VALEURS];
long long Occ[MAX_VALEURS];
long long DP[MAX_VALEURS][MAX_VALEURS];
long long nbNoeuds, taille;
void Dfs(long long noeud, long long pere, long long dist) {
  Occ[dist]++;
  for (long long dest : Voisins[noeud]) {
    if (dest != pere) Dfs(dest, noeud, dist + 1);
  }
  return;
}
void Solve() {
  scanf("%lld %lld", &nbNoeuds, &taille);
  for (long long i = 0; i < nbNoeuds; i++) {
    Voisins[i].clear();
  }
  for (long long i = 1; i < nbNoeuds; i++) {
    long long a, b;
    scanf("%lld %lld", &a, &b);
    Voisins[--a].push_back(--b);
    Voisins[b].push_back(a);
  }
  if (taille == 2) {
    printf("%lld\n", (nbNoeuds * (nbNoeuds - 1)) / 2);
    return;
  }
  long long ans = 0;
  for (long long i = 0; i < nbNoeuds; i++) {
    for (long long j = 0; j < nbNoeuds; j++) {
      fill_n(DP[j], nbNoeuds, 0LL);
      DP[j][0] = 1;
    }
    for (long long dest : Voisins[i]) {
      fill_n(Occ, nbNoeuds, 0LL);
      Dfs(dest, i, 1);
      for (long long d = 0; d < nbNoeuds; d++) {
        for (long long j = nbNoeuds - 1; j > 0; j--) {
          DP[d][j] += DP[d][j - 1] * Occ[d];
          DP[d][j] %= MOD;
        }
      }
    }
    for (long long d = 0; d < nbNoeuds; d++) {
      ans += DP[d][taille];
    }
    ans %= MOD;
  }
  printf("%lld\n", ans);
  return;
}
signed main() {
  long long nbTests = 1;
  scanf("%lld", &nbTests);
  while (nbTests--) {
    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 MAX_VALEURS = (101);
vector<long long> Voisins[MAX_VALEURS];
long long Occ[MAX_VALEURS];
long long DP[MAX_VALEURS][MAX_VALEURS];
long long nbNoeuds, taille;
void Dfs(long long noeud, long long pere, long long dist) {
  Occ[dist]++;
  for (long long dest : Voisins[noeud]) {
    if (dest != pere) Dfs(dest, noeud, dist + 1);
  }
  return;
}
void Solve() {
  scanf("%lld %lld", &nbNoeuds, &taille);
  for (long long i = 0; i < nbNoeuds; i++) {
    Voisins[i].clear();
  }
  for (long long i = 1; i < nbNoeuds; i++) {
    long long a, b;
    scanf("%lld %lld", &a, &b);
    Voisins[--a].push_back(--b);
    Voisins[b].push_back(a);
  }
  if (taille == 2) {
    printf("%lld\n", (nbNoeuds * (nbNoeuds - 1)) / 2);
    return;
  }
  long long ans = 0;
  for (long long i = 0; i < nbNoeuds; i++) {
    for (long long j = 0; j < nbNoeuds; j++) {
      fill_n(DP[j], nbNoeuds, 0LL);
      DP[j][0] = 1;
    }
    for (long long dest : Voisins[i]) {
      fill_n(Occ, nbNoeuds, 0LL);
      Dfs(dest, i, 1);
      for (long long d = 0; d < nbNoeuds; d++) {
        for (long long j = nbNoeuds - 1; j > 0; j--) {
          DP[d][j] += DP[d][j - 1] * Occ[d];
          DP[d][j] %= MOD;
        }
      }
    }
    for (long long d = 0; d < nbNoeuds; d++) {
      ans += DP[d][taille];
    }
    ans %= MOD;
  }
  printf("%lld\n", ans);
  return;
}
signed main() {
  long long nbTests = 1;
  scanf("%lld", &nbTests);
  while (nbTests--) {
    Solve();
  }
  return 0;
}

```