#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vii = vector<pii>;
using vll = vector<pll>;
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
vector<int> sieve;
void calc_sieve(int n) {
  sieve.reserve(n + 1);
  for (int i = 0, nnnn = (n + 1); i < nnnn; ++i) sieve[i] = i;
  for (int i = 2, nnnn = (n + 1); i < nnnn; ++i) {
    if (sieve[i] == i) {
      for (ll j = i * i; j <= n; j += i) {
        if (sieve[j] == j) sieve[j] = i;
      }
    }
  }
}
ll norm(ll a, ll P) {
  while (a >= P) a -= P;
  while (a < 0) a += P;
  return a;
}
ll mul_mod(ll a, ll b, ll P) { return a * b % P; }
ll pow_mod(ll a, ll b, ll P) {
  ll curr = a, ans = 1;
  while (b != 0) {
    if (b & 1) {
      ans *= curr;
      ans %= P;
    }
    curr *= curr;
    curr %= P;
    b >>= 1;
  }
  return ans;
}
ll inv_mod(ll a, ll P) { return pow_mod(a, P - 2ll, P); }
ll div_mod(ll a, ll b, ll P) { return a * inv_mod(b, P) % P; }
int msb(ll a) {
  for (int i = 62; i >= 0; i--) {
    if (a & ((ll)1 << i)) return i + 1;
  }
  return 0;
}
vector<ll> fact;
void calc_fact(int n, ll P) {
  if (fact.size() < 1) fact.push_back(1);
  for (int i = fact.size(), nnnn = (n + 1); i < nnnn; ++i)
    fact.push_back(fact[i - 1] * i % P);
}
vector<ll> fact_inv;
void calc_fact_inv(int n, ll P) {
  if (fact_inv.size() < 1) fact_inv.push_back(1);
  for (int i = fact_inv.size(), nnnn = (n + 1); i < nnnn; ++i)
    fact_inv.push_back(fact_inv[i - 1] * inv_mod(i, P) % P);
}
ll nCr(int n, int r, ll P) {
  if (r > n) return 0;
  calc_fact(n, P);
  calc_fact_inv(n, P);
  return (fact[n] * fact_inv[r] % P) * fact_inv[n - r] % P;
}
ll nCr_inv(int n, int r, ll P) {
  if (r > n) return 0;
  calc_fact(n, P);
  calc_fact_inv(n, P);
  return (fact_inv[n] * fact[r] % P) * fact[n - r] % P;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long double eps = 1e-9;
const long long mod = 1000000007;
const int MAXN = 100;
vector<int> adjlist[MAXN];
vector<vector<int>> v;
int n, k;
ll f[MAXN], invf[MAXN];
void dfs(int curr, int dist, int p, vector<int> &v) {
  v[dist]++;
  for (int i : adjlist[curr]) {
    if (i != p) {
      dfs(i, dist + 1, curr, v);
    }
  }
}
void solve() {
  cin >> n >> k;
  for (int i = 0, nnnn = (n); i < nnnn; ++i) adjlist[i].clear();
  for (int i = 0, nnnn = (n - 1); i < nnnn; ++i) {
    int a, b;
    cin >> a >> b;
    adjlist[a - 1].push_back(b - 1);
    adjlist[b - 1].push_back(a - 1);
  }
  if (k == 2) {
    cout << (n - 1) * n / 2 << endl;
  } else {
    ll ans = 0;
    for (int i = 0, nnnn = (n); i < nnnn; ++i) {
      v = vector<vector<int>>(adjlist[i].size(), vector<int>(n, 0));
      for (int j = 0, nnnn = (adjlist[i].size()); j < nnnn; ++j) {
        dfs(adjlist[i][j], 0, i, v[j]);
      }
      for (int j = 0, nnnn = (n); j < nnnn; ++j) {
        ll dp[k];
        memset(dp, 0, sizeof dp);
        for (int l = 0, nnnn = (adjlist[i].size()); l < nnnn; ++l) {
          for (int m = k - 1; m >= 1; m--)
            dp[m] += dp[m - 1] * v[l][j], dp[m] %= mod;
          dp[0] += v[l][j];
        }
        ans += dp[k - 1];
      }
    }
    cout << ans % mod << endl;
  }
}
int main() {
  f[0] = 1;
  invf[0] = 1;
  for (int i = 1, nnnn = (MAXN); i < nnnn; ++i)
    f[i] = f[i - 1] * i % mod, invf[i] = invf[i - 1] * inv_mod(i, mod) % mod;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(12) << fixed;
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Your challenge is to write 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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vii = vector<pii>;
using vll = vector<pll>;
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
vector<int> sieve;
void calc_sieve(int n) {
  sieve.reserve(n + 1);
  for (int i = 0, nnnn = (n + 1); i < nnnn; ++i) sieve[i] = i;
  for (int i = 2, nnnn = (n + 1); i < nnnn; ++i) {
    if (sieve[i] == i) {
      for (ll j = i * i; j <= n; j += i) {
        if (sieve[j] == j) sieve[j] = i;
      }
    }
  }
}
ll norm(ll a, ll P) {
  while (a >= P) a -= P;
  while (a < 0) a += P;
  return a;
}
ll mul_mod(ll a, ll b, ll P) { return a * b % P; }
ll pow_mod(ll a, ll b, ll P) {
  ll curr = a, ans = 1;
  while (b != 0) {
    if (b & 1) {
      ans *= curr;
      ans %= P;
    }
    curr *= curr;
    curr %= P;
    b >>= 1;
  }
  return ans;
}
ll inv_mod(ll a, ll P) { return pow_mod(a, P - 2ll, P); }
ll div_mod(ll a, ll b, ll P) { return a * inv_mod(b, P) % P; }
int msb(ll a) {
  for (int i = 62; i >= 0; i--) {
    if (a & ((ll)1 << i)) return i + 1;
  }
  return 0;
}
vector<ll> fact;
void calc_fact(int n, ll P) {
  if (fact.size() < 1) fact.push_back(1);
  for (int i = fact.size(), nnnn = (n + 1); i < nnnn; ++i)
    fact.push_back(fact[i - 1] * i % P);
}
vector<ll> fact_inv;
void calc_fact_inv(int n, ll P) {
  if (fact_inv.size() < 1) fact_inv.push_back(1);
  for (int i = fact_inv.size(), nnnn = (n + 1); i < nnnn; ++i)
    fact_inv.push_back(fact_inv[i - 1] * inv_mod(i, P) % P);
}
ll nCr(int n, int r, ll P) {
  if (r > n) return 0;
  calc_fact(n, P);
  calc_fact_inv(n, P);
  return (fact[n] * fact_inv[r] % P) * fact_inv[n - r] % P;
}
ll nCr_inv(int n, int r, ll P) {
  if (r > n) return 0;
  calc_fact(n, P);
  calc_fact_inv(n, P);
  return (fact_inv[n] * fact[r] % P) * fact[n - r] % P;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long double eps = 1e-9;
const long long mod = 1000000007;
const int MAXN = 100;
vector<int> adjlist[MAXN];
vector<vector<int>> v;
int n, k;
ll f[MAXN], invf[MAXN];
void dfs(int curr, int dist, int p, vector<int> &v) {
  v[dist]++;
  for (int i : adjlist[curr]) {
    if (i != p) {
      dfs(i, dist + 1, curr, v);
    }
  }
}
void solve() {
  cin >> n >> k;
  for (int i = 0, nnnn = (n); i < nnnn; ++i) adjlist[i].clear();
  for (int i = 0, nnnn = (n - 1); i < nnnn; ++i) {
    int a, b;
    cin >> a >> b;
    adjlist[a - 1].push_back(b - 1);
    adjlist[b - 1].push_back(a - 1);
  }
  if (k == 2) {
    cout << (n - 1) * n / 2 << endl;
  } else {
    ll ans = 0;
    for (int i = 0, nnnn = (n); i < nnnn; ++i) {
      v = vector<vector<int>>(adjlist[i].size(), vector<int>(n, 0));
      for (int j = 0, nnnn = (adjlist[i].size()); j < nnnn; ++j) {
        dfs(adjlist[i][j], 0, i, v[j]);
      }
      for (int j = 0, nnnn = (n); j < nnnn; ++j) {
        ll dp[k];
        memset(dp, 0, sizeof dp);
        for (int l = 0, nnnn = (adjlist[i].size()); l < nnnn; ++l) {
          for (int m = k - 1; m >= 1; m--)
            dp[m] += dp[m - 1] * v[l][j], dp[m] %= mod;
          dp[0] += v[l][j];
        }
        ans += dp[k - 1];
      }
    }
    cout << ans % mod << endl;
  }
}
int main() {
  f[0] = 1;
  invf[0] = 1;
  for (int i = 1, nnnn = (MAXN); i < nnnn; ++i)
    f[i] = f[i - 1] * i % mod, invf[i] = invf[i - 1] * inv_mod(i, mod) % mod;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(12) << fixed;
  int t;
  cin >> t;
  while (t--) solve();
}

```