#include <bits/stdc++.h>
using namespace std;
void main0();
int main() {
  clock_t start, end;
  ios::sync_with_stdio(false);
  cin.tie(0);
  main0();
  return 0;
}
const int dx[8] = {0, 1, -1, 0, 1, 1, -1, -1};
const int dy[8] = {1, 0, 0, -1, 1, -1, -1, 1};
const int N = 105;
const int M = 1e5;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-6;
const double Pi = acos(-1.0);
mt19937 rnd(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
long long n, k;
long long ans;
vector<int> e[N];
int deg[N];
int dep[N][N];
int cnt[N][N];
void dfs(int rt, int u, int fa) {
  dep[rt][u] = dep[rt][fa] + 1;
  cnt[rt][dep[rt][u]]++;
  for (int v : e[u])
    if (v != fa) {
      dfs(rt, v, u);
    }
}
long long dp[N][N];
void check(int u) {
  for (int v : e[u]) {
    dep[v][u] = 0;
    for (int i = 1; i <= n; ++i) {
      cnt[v][i] = 0;
    }
    dfs(v, v, u);
  }
  for (int d = 1; d <= n; ++d) {
    memset(dp, 0, sizeof dp);
    dp[0][0] = 1;
    for (int i = 1; i <= e[u].size(); ++i) {
      for (int j = 0; j <= k; ++j) {
        dp[i][j] += dp[i - 1][j];
      }
      for (int j = 1; j <= k; ++j) {
        dp[i][j] += dp[i - 1][j - 1] * cnt[e[u][i - 1]][d] % mod;
        dp[i][j] %= mod;
      }
    }
    ans += dp[e[u].size()][k];
    ans %= mod;
  }
}
void main0() {
  int T;
  cin >> T;
  while (T--) {
    ans = 0;
    cin >> n >> k;
    for (int i = 1; i <= n; ++i) {
      e[i].clear();
      deg[i] = 0;
    }
    for (int i = 1; i <= n - 1; ++i) {
      int u, v;
      cin >> u >> v;
      deg[u]++, deg[v]++;
      e[u].push_back(v);
      e[v].push_back(u);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << endl;
      continue;
    }
    for (int i = 1; i <= n; ++i) {
      if (deg[i] >= k) check(i);
    }
    cout << ans << endl;
  }
}

### 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;
void main0();
int main() {
  clock_t start, end;
  ios::sync_with_stdio(false);
  cin.tie(0);
  main0();
  return 0;
}
const int dx[8] = {0, 1, -1, 0, 1, 1, -1, -1};
const int dy[8] = {1, 0, 0, -1, 1, -1, -1, 1};
const int N = 105;
const int M = 1e5;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-6;
const double Pi = acos(-1.0);
mt19937 rnd(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
long long n, k;
long long ans;
vector<int> e[N];
int deg[N];
int dep[N][N];
int cnt[N][N];
void dfs(int rt, int u, int fa) {
  dep[rt][u] = dep[rt][fa] + 1;
  cnt[rt][dep[rt][u]]++;
  for (int v : e[u])
    if (v != fa) {
      dfs(rt, v, u);
    }
}
long long dp[N][N];
void check(int u) {
  for (int v : e[u]) {
    dep[v][u] = 0;
    for (int i = 1; i <= n; ++i) {
      cnt[v][i] = 0;
    }
    dfs(v, v, u);
  }
  for (int d = 1; d <= n; ++d) {
    memset(dp, 0, sizeof dp);
    dp[0][0] = 1;
    for (int i = 1; i <= e[u].size(); ++i) {
      for (int j = 0; j <= k; ++j) {
        dp[i][j] += dp[i - 1][j];
      }
      for (int j = 1; j <= k; ++j) {
        dp[i][j] += dp[i - 1][j - 1] * cnt[e[u][i - 1]][d] % mod;
        dp[i][j] %= mod;
      }
    }
    ans += dp[e[u].size()][k];
    ans %= mod;
  }
}
void main0() {
  int T;
  cin >> T;
  while (T--) {
    ans = 0;
    cin >> n >> k;
    for (int i = 1; i <= n; ++i) {
      e[i].clear();
      deg[i] = 0;
    }
    for (int i = 1; i <= n - 1; ++i) {
      int u, v;
      cin >> u >> v;
      deg[u]++, deg[v]++;
      e[u].push_back(v);
      e[v].push_back(u);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << endl;
      continue;
    }
    for (int i = 1; i <= n; ++i) {
      if (deg[i] >= k) check(i);
    }
    cout << ans << endl;
  }
}

```