#include <bits/stdc++.h>
using namespace std;
vector<string> vec_splitter(string s) {
  for (char& c : s) c = c == ',' ? ' ' : c;
  stringstream ss;
  ss << s;
  vector<string> res;
  for (string z; ss >> z; res.push_back(z))
    ;
  return res;
}
void debug_out(vector<string> args, int idx) { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, Head H, Tail... T) {
  if (idx > 0) cerr << ", ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, T...);
}
void localTest() {}
const long long MOD = 1e9 + 7, N = 105;
vector<long long> V[N];
long long dp[N][N], level[N];
void dfs(long long node, long long p, long long l) {
  level[l]++;
  for (auto child : V[node]) {
    if (child == p) {
      continue;
    }
    dfs(child, node, l + 1);
  }
  return;
}
void add(long long& x, long long y) {
  x += y;
  x %= MOD;
  return;
}
long long mul(long long x, long long y) {
  x %= MOD;
  y %= MOD;
  return (x * y) % MOD;
}
void solver() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    V[i].clear();
  }
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    V[x].push_back(y);
    V[y].push_back(x);
  }
  if (k == 2) {
    cout << (n * (n - 1)) / 2 << "\n";
    return;
  }
  long long ans = 0;
  for (int root = 1; root <= n; ++root) {
    if ((int)V[root].size() < k) {
      continue;
    }
    memset(dp, 0, sizeof dp);
    for (int i = 1; i <= n; ++i) {
      dp[i][0] = 1;
    }
    for (auto u : V[root]) {
      memset(level, 0, sizeof level);
      dfs(u, root, 1);
      for (int i = 1; i <= n; ++i) {
        for (int j = k - 1; j >= 0; --j) {
          add(dp[i][j + 1], mul(dp[i][j], level[i]));
        }
      }
    }
    for (int i = 1; i <= n; ++i) {
      add(ans, dp[i][k]);
    }
  }
  cout << ans << "\n";
  return;
}
int main() {
  localTest();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solver();
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
vector<string> vec_splitter(string s) {
  for (char& c : s) c = c == ',' ? ' ' : c;
  stringstream ss;
  ss << s;
  vector<string> res;
  for (string z; ss >> z; res.push_back(z))
    ;
  return res;
}
void debug_out(vector<string> args, int idx) { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, Head H, Tail... T) {
  if (idx > 0) cerr << ", ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, T...);
}
void localTest() {}
const long long MOD = 1e9 + 7, N = 105;
vector<long long> V[N];
long long dp[N][N], level[N];
void dfs(long long node, long long p, long long l) {
  level[l]++;
  for (auto child : V[node]) {
    if (child == p) {
      continue;
    }
    dfs(child, node, l + 1);
  }
  return;
}
void add(long long& x, long long y) {
  x += y;
  x %= MOD;
  return;
}
long long mul(long long x, long long y) {
  x %= MOD;
  y %= MOD;
  return (x * y) % MOD;
}
void solver() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    V[i].clear();
  }
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    V[x].push_back(y);
    V[y].push_back(x);
  }
  if (k == 2) {
    cout << (n * (n - 1)) / 2 << "\n";
    return;
  }
  long long ans = 0;
  for (int root = 1; root <= n; ++root) {
    if ((int)V[root].size() < k) {
      continue;
    }
    memset(dp, 0, sizeof dp);
    for (int i = 1; i <= n; ++i) {
      dp[i][0] = 1;
    }
    for (auto u : V[root]) {
      memset(level, 0, sizeof level);
      dfs(u, root, 1);
      for (int i = 1; i <= n; ++i) {
        for (int j = k - 1; j >= 0; --j) {
          add(dp[i][j + 1], mul(dp[i][j], level[i]));
        }
      }
    }
    for (int i = 1; i <= n; ++i) {
      add(ans, dp[i][k]);
    }
  }
  cout << ans << "\n";
  return;
}
int main() {
  localTest();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solver();
  }
  return 0;
}

```