#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const long long oo = 1e18;
const long double eps = 1e-9;
const string yes = "YES";
const string no = "NO";
const long double PI = acos(-1.0);
void io(long long x) {
  if (!x) {
    freopen(
        "1551f"
        ".inp",
        "r", stdin);
    return;
  }
  if (x == 1) {
    freopen(
        "1551f"
        ".inp",
        "r", stdin);
    freopen(
        "1551f"
        ".out",
        "w", stdout);
    return;
  }
  freopen(
      "1551f"
      ".out",
      "w", stdout);
}
long long n, k;
const long long N = 105;
vector<long long> g[N];
long long dp[N][N];
vector<long long> cnt(N);
const long long MOD = 1e9 + 7;
long long fast_pow(long long a, long long b) {
  long long res = 1;
  for (; b; b /= 2) {
    if (b & 1) {
      ((res *= a) %= MOD);
      b--;
    }
    ((a *= a) %= MOD);
  }
  return res;
}
long long inv(long long x) { return fast_pow(x, MOD - 2); }
void add(long long &a, long long b) { (a += b) %= MOD; }
void sub(long long &a, long long b) { (a -= b - MOD) %= MOD; }
void mul(long long &a, long long b) { (a *= b) %= MOD; }
void div(long long &a, long long b) { (a *= inv(b)) %= MOD; }
void init(long long x) {
  for (long long i = 0; i <= x; i++) {
    g[i].clear();
  }
}
void dfs(long long u, long long pre, long long d, vector<long long> &cnt) {
  cnt[d + 1]++;
  for (long long v : g[u]) {
    if (v ^ pre) dfs(v, u, d + 1, cnt);
  }
}
void solve() {
  cin >> n >> k;
  init(n);
  for (long long i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  if (k == 2) {
    cout << (n * (n - 1)) / 2 << '\n';
    return;
  }
  long long ans = 0;
  for (long long u = 1; u <= n; u++) {
    for (long long i = 0; i <= n; i++) {
      for (long long j = 0; j <= n; j++) {
        dp[i][j] = 0;
        dp[0][j] = 1;
      }
    }
    for (long long v : g[u]) {
      for (long long i = 0; i <= n; i++) {
        cnt[i] = 0;
      }
      dfs(v, u, 0, cnt);
      for (long long i = k; i; i--) {
        for (long long d = 1; d <= n; d++) {
          add(dp[i][d], (dp[i - 1][d] * cnt[d]) % MOD);
        }
      }
    }
    for (long long d = 1; d <= n; d++) {
      add(ans, dp[k][d]);
    }
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long test_case = 1;
  cin >> test_case;
  for (long long tt = 1; tt <= test_case; tt++) {
    solve();
  }
  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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const long long oo = 1e18;
const long double eps = 1e-9;
const string yes = "YES";
const string no = "NO";
const long double PI = acos(-1.0);
void io(long long x) {
  if (!x) {
    freopen(
        "1551f"
        ".inp",
        "r", stdin);
    return;
  }
  if (x == 1) {
    freopen(
        "1551f"
        ".inp",
        "r", stdin);
    freopen(
        "1551f"
        ".out",
        "w", stdout);
    return;
  }
  freopen(
      "1551f"
      ".out",
      "w", stdout);
}
long long n, k;
const long long N = 105;
vector<long long> g[N];
long long dp[N][N];
vector<long long> cnt(N);
const long long MOD = 1e9 + 7;
long long fast_pow(long long a, long long b) {
  long long res = 1;
  for (; b; b /= 2) {
    if (b & 1) {
      ((res *= a) %= MOD);
      b--;
    }
    ((a *= a) %= MOD);
  }
  return res;
}
long long inv(long long x) { return fast_pow(x, MOD - 2); }
void add(long long &a, long long b) { (a += b) %= MOD; }
void sub(long long &a, long long b) { (a -= b - MOD) %= MOD; }
void mul(long long &a, long long b) { (a *= b) %= MOD; }
void div(long long &a, long long b) { (a *= inv(b)) %= MOD; }
void init(long long x) {
  for (long long i = 0; i <= x; i++) {
    g[i].clear();
  }
}
void dfs(long long u, long long pre, long long d, vector<long long> &cnt) {
  cnt[d + 1]++;
  for (long long v : g[u]) {
    if (v ^ pre) dfs(v, u, d + 1, cnt);
  }
}
void solve() {
  cin >> n >> k;
  init(n);
  for (long long i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  if (k == 2) {
    cout << (n * (n - 1)) / 2 << '\n';
    return;
  }
  long long ans = 0;
  for (long long u = 1; u <= n; u++) {
    for (long long i = 0; i <= n; i++) {
      for (long long j = 0; j <= n; j++) {
        dp[i][j] = 0;
        dp[0][j] = 1;
      }
    }
    for (long long v : g[u]) {
      for (long long i = 0; i <= n; i++) {
        cnt[i] = 0;
      }
      dfs(v, u, 0, cnt);
      for (long long i = k; i; i--) {
        for (long long d = 1; d <= n; d++) {
          add(dp[i][d], (dp[i - 1][d] * cnt[d]) % MOD);
        }
      }
    }
    for (long long d = 1; d <= n; d++) {
      add(ans, dp[k][d]);
    }
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long test_case = 1;
  cin >> test_case;
  for (long long tt = 1; tt <= test_case; tt++) {
    solve();
  }
  return 0;
}

```