#include <bits/stdc++.h>
double eps = 1e-9;
using namespace std;
vector<vector<int> > v;
int parent[50002];
int disc[50002];
int dp[50002][502];
int dp1[50002][502];
void dfs(int u, int p) {
  disc[u] = 0;
  for (int i = 0; i < v[u].size(); i++) {
    int front = v[u][i];
    if (front == p) {
      continue;
    }
    if (disc[front] == -1) {
      parent[front] = u;
      dfs(front, u);
    }
  }
}
int main() {
  int n, z;
  cin >> n >> z;
  v.resize(n + 1);
  int a, b;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d", &a);
    scanf("%d", &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  memset(parent, -1, sizeof(parent));
  memset(disc, -1, sizeof(disc));
  dfs(1, -1);
  for (int i = 1; i <= n; i++) {
    dp[i][0] = 1;
  }
  long long sum = 0;
  long long sp = 0;
  for (int i = 1; i <= z; i++) {
    for (int j = 1; j <= n; j++) {
      for (int k = 0; k < v[j].size(); k++) {
        int front = v[j][k];
        if (front != parent[j]) {
          dp[j][i] += (dp[front][i - 1]);
        } else if (i > 1) {
          dp1[j][i] += (dp[front][i - 1] - dp[j][i - 2] + dp1[front][i - 1]);
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    sum += dp[i][z];
    sp += dp1[i][z];
  }
  cout << (sp / 2 + sum);
  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>
double eps = 1e-9;
using namespace std;
vector<vector<int> > v;
int parent[50002];
int disc[50002];
int dp[50002][502];
int dp1[50002][502];
void dfs(int u, int p) {
  disc[u] = 0;
  for (int i = 0; i < v[u].size(); i++) {
    int front = v[u][i];
    if (front == p) {
      continue;
    }
    if (disc[front] == -1) {
      parent[front] = u;
      dfs(front, u);
    }
  }
}
int main() {
  int n, z;
  cin >> n >> z;
  v.resize(n + 1);
  int a, b;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d", &a);
    scanf("%d", &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  memset(parent, -1, sizeof(parent));
  memset(disc, -1, sizeof(disc));
  dfs(1, -1);
  for (int i = 1; i <= n; i++) {
    dp[i][0] = 1;
  }
  long long sum = 0;
  long long sp = 0;
  for (int i = 1; i <= z; i++) {
    for (int j = 1; j <= n; j++) {
      for (int k = 0; k < v[j].size(); k++) {
        int front = v[j][k];
        if (front != parent[j]) {
          dp[j][i] += (dp[front][i - 1]);
        } else if (i > 1) {
          dp1[j][i] += (dp[front][i - 1] - dp[j][i - 2] + dp1[front][i - 1]);
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    sum += dp[i][z];
    sp += dp1[i][z];
  }
  cout << (sp / 2 + sum);
  return 0;
}

```