#include <bits/stdc++.h>
using namespace std;
const int INF = (int)2e9;
const double EPS = 1e-6;
const int MOD = 1000003;
vector<vector<int> > g(50000);
int n, k, dp[50000][501];
long long ans;
void dfs(int v, int p = -1) {
  dp[v][0] = 1;
  for (int i = 0; i < g[v].size(); ++i) {
    int u = g[v][i];
    if (u == p) continue;
    dfs(u, v);
    for (int j = 1; j < k; ++j) ans += dp[v][j] * dp[u][k - j - 1];
    for (int j = 1; j <= k; ++j) dp[v][j] += dp[u][j - 1];
  }
  ans += dp[v][k];
}
int main() {
  int a, b;
  cin >> n >> k;
  for (int i = 0; i < n - 1; ++i) {
    cin >> a >> b;
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0);
  cout << ans;
  return 0;
}

### Prompt

Generate a Cpp solution 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;
const int INF = (int)2e9;
const double EPS = 1e-6;
const int MOD = 1000003;
vector<vector<int> > g(50000);
int n, k, dp[50000][501];
long long ans;
void dfs(int v, int p = -1) {
  dp[v][0] = 1;
  for (int i = 0; i < g[v].size(); ++i) {
    int u = g[v][i];
    if (u == p) continue;
    dfs(u, v);
    for (int j = 1; j < k; ++j) ans += dp[v][j] * dp[u][k - j - 1];
    for (int j = 1; j <= k; ++j) dp[v][j] += dp[u][j - 1];
  }
  ans += dp[v][k];
}
int main() {
  int a, b;
  cin >> n >> k;
  for (int i = 0; i < n - 1; ++i) {
    cin >> a >> b;
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0);
  cout << ans;
  return 0;
}

```