#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
const ll mod = 1e9 + 7;
const ll N = 1e5 + 5;
ll n, k;
vector<ll> a[50005];
ll kq = 0;
ll dp[50005][505];
ll par[50005];
void dfs(ll u) {
  dp[u][0] = 1;
  for (auto x : a[u])
    if (x != par[u]) {
      par[x] = u;
      dp[u][1]++;
      dfs(x);
      for (ll i = 1; i <= k - 1; i++) dp[u][i + 1] += dp[x][i];
    }
}
int32_t main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (ll i = 1; i <= n - 1; i++) {
    ll x, y;
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  for (ll i = 1; i <= n; i++) par[i] = i;
  dfs(1);
  for (ll i = 1; i <= n; i++) {
    kq += dp[i][k];
  }
  for (ll i = 1; i <= n; i++)
    for (ll j = 1; j <= k / 2; j++) {
      {
        ll tmp = 0;
        for (auto x : a[i])
          if (x != par[i]) {
            tmp += (dp[x][j - 1]) * (dp[i][k - j] - dp[x][k - j - 1]);
          }
        if (j + j == k) tmp /= 2;
        kq += tmp;
      }
    }
  cout << kq;
}

### Prompt

In cpp, your task is to solve the following problem:

A tree is a connected graph that doesn't contain any cycles.

The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.

You are given a tree with n vertices and a positive number k. Find the number of distinct pairs of the vertices which have a distance of exactly k between them. Note that pairs (v, u) and (u, v) are considered to be the same pair.

Input

The first line contains two integers n and k (1 ≤ n ≤ 50000, 1 ≤ k ≤ 500) — the number of vertices and the required distance between the vertices.

Next n - 1 lines describe the edges as "ai bi" (without the quotes) (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the vertices connected by the i-th edge. All given edges are different.

Output

Print a single integer — the number of distinct pairs of the tree's vertices which have a distance of exactly k between them.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

5 2
1 2
2 3
3 4
2 5


Output

4


Input

5 3
1 2
2 3
3 4
4 5


Output

2

Note

In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
const ll mod = 1e9 + 7;
const ll N = 1e5 + 5;
ll n, k;
vector<ll> a[50005];
ll kq = 0;
ll dp[50005][505];
ll par[50005];
void dfs(ll u) {
  dp[u][0] = 1;
  for (auto x : a[u])
    if (x != par[u]) {
      par[x] = u;
      dp[u][1]++;
      dfs(x);
      for (ll i = 1; i <= k - 1; i++) dp[u][i + 1] += dp[x][i];
    }
}
int32_t main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (ll i = 1; i <= n - 1; i++) {
    ll x, y;
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  for (ll i = 1; i <= n; i++) par[i] = i;
  dfs(1);
  for (ll i = 1; i <= n; i++) {
    kq += dp[i][k];
  }
  for (ll i = 1; i <= n; i++)
    for (ll j = 1; j <= k / 2; j++) {
      {
        ll tmp = 0;
        for (auto x : a[i])
          if (x != par[i]) {
            tmp += (dp[x][j - 1]) * (dp[i][k - j] - dp[x][k - j - 1]);
          }
        if (j + j == k) tmp /= 2;
        kq += tmp;
      }
    }
  cout << kq;
}

```