#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void checkmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void checkmax(T &a, T b) {
  if (b > a) a = b;
}
int N, K;
vector<int> G[50061];
int dp[50061][510];
long long ans;
void dfs(int u, int p) {
  for (int i = (0); i < (int)(((int)((G[u]).size()))); i++) {
    int v = G[u][i];
    if (v == p) continue;
    dfs(v, u);
    for (int j = (0); j < (int)(K); j++) dp[u][j + 1] += dp[v][j];
  }
  dp[u][0] = 1;
}
void rec(int u, int p) {
  ans += dp[u][K];
  if (p != -1) {
    ans += dp[p][K - 1];
    if (K - 2 >= 0) ans -= dp[u][K - 2];
    for (int j = ((K)-1); j >= (int)(0); j--) {
      dp[u][j + 1] += dp[p][j];
      if (j > 0) dp[u][j + 1] -= dp[u][j - 1];
    }
  }
  for (int i = (0); i < (int)(((int)((G[u]).size()))); i++) {
    int v = G[u][i];
    if (v == p) continue;
    rec(v, u);
  }
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = (0); i < (int)(N - 1); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--, v--;
    G[u].push_back(v), G[v].push_back(u);
  }
  memset((dp), 0, sizeof(dp));
  dfs(0, -1);
  ans = 0;
  rec(0, -1);
  ans /= 2;
  cout << ans << endl;
  return 0;
}

### 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;
template <class T>
inline void checkmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void checkmax(T &a, T b) {
  if (b > a) a = b;
}
int N, K;
vector<int> G[50061];
int dp[50061][510];
long long ans;
void dfs(int u, int p) {
  for (int i = (0); i < (int)(((int)((G[u]).size()))); i++) {
    int v = G[u][i];
    if (v == p) continue;
    dfs(v, u);
    for (int j = (0); j < (int)(K); j++) dp[u][j + 1] += dp[v][j];
  }
  dp[u][0] = 1;
}
void rec(int u, int p) {
  ans += dp[u][K];
  if (p != -1) {
    ans += dp[p][K - 1];
    if (K - 2 >= 0) ans -= dp[u][K - 2];
    for (int j = ((K)-1); j >= (int)(0); j--) {
      dp[u][j + 1] += dp[p][j];
      if (j > 0) dp[u][j + 1] -= dp[u][j - 1];
    }
  }
  for (int i = (0); i < (int)(((int)((G[u]).size()))); i++) {
    int v = G[u][i];
    if (v == p) continue;
    rec(v, u);
  }
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = (0); i < (int)(N - 1); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--, v--;
    G[u].push_back(v), G[v].push_back(u);
  }
  memset((dp), 0, sizeof(dp));
  dfs(0, -1);
  ans = 0;
  rec(0, -1);
  ans /= 2;
  cout << ans << endl;
  return 0;
}

```