#include <bits/stdc++.h>
using namespace std;
void fast() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const long long MAXN = 5e4 + 5;
const long long k = 505;
long long dp[MAXN][k];
vector<long long> edg[MAXN];
long long ans = 0, ka, sumyd = 0;
void dfs(long long v, long long p) {
  dp[v][0] = 1;
  long long i, a, b, x, y;
  for (auto u : edg[v])
    if (u != p) {
      dfs(u, v);
      for (i = 0; i < k - 2; i++) dp[v][i + 1] += dp[u][i];
    }
  for (auto u : edg[v]) {
    if (u == p) continue;
    for (i = 1; i * 2 <= ka; i++) {
      a = i;
      b = ka - i;
      x = dp[u][a - 1];
      y = dp[v][b] - dp[u][b - 1];
      if (i * 2 == ka)
        sumyd += x * y;
      else
        ans += x * y;
    }
  }
  ans += dp[v][ka];
}
signed main() {
  long long n, i, a, b, j;
  cin >> n >> ka;
  for (i = 0; i < n - 1; i++) {
    cin >> a >> b;
    edg[a].push_back(b);
    edg[b].push_back(a);
  }
  dfs(1, 0);
  cout << ans + sumyd / 2 << '\n';
  return 0;
}

### Prompt

Please create a solution in CPP to 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;
void fast() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const long long MAXN = 5e4 + 5;
const long long k = 505;
long long dp[MAXN][k];
vector<long long> edg[MAXN];
long long ans = 0, ka, sumyd = 0;
void dfs(long long v, long long p) {
  dp[v][0] = 1;
  long long i, a, b, x, y;
  for (auto u : edg[v])
    if (u != p) {
      dfs(u, v);
      for (i = 0; i < k - 2; i++) dp[v][i + 1] += dp[u][i];
    }
  for (auto u : edg[v]) {
    if (u == p) continue;
    for (i = 1; i * 2 <= ka; i++) {
      a = i;
      b = ka - i;
      x = dp[u][a - 1];
      y = dp[v][b] - dp[u][b - 1];
      if (i * 2 == ka)
        sumyd += x * y;
      else
        ans += x * y;
    }
  }
  ans += dp[v][ka];
}
signed main() {
  long long n, i, a, b, j;
  cin >> n >> ka;
  for (i = 0; i < n - 1; i++) {
    cin >> a >> b;
    edg[a].push_back(b);
    edg[b].push_back(a);
  }
  dfs(1, 0);
  cout << ans + sumyd / 2 << '\n';
  return 0;
}

```