#include <bits/stdc++.h>
using namespace std;
void init_cin() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
}
int d[55555][555];
vector<int> g[55555];
long long ans = 0;
int n, k;
void dfs(int v, int p) {
  for (int i : g[v]) {
    if (i == p) {
      continue;
    }
    dfs(i, v);
    for (int j = 1; j <= k; j++) {
      d[v][j] += d[i][j - 1];
    }
  }
  ans += d[v][k];
  long long t = 0;
  for (int i : g[v]) {
    if (i == p) {
      continue;
    }
    for (int j = 0; j + 2 <= k; j++) {
      t += 1LL * d[i][j] * (d[v][k - j - 1] - d[i][k - j - 2]);
    }
  }
  t /= 2;
  ans += t;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a;
    --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int i = 0; i < n; i++) {
    d[i][0] = 1;
  }
  dfs(0, -1);
  cout << ans;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 init_cin() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
}
int d[55555][555];
vector<int> g[55555];
long long ans = 0;
int n, k;
void dfs(int v, int p) {
  for (int i : g[v]) {
    if (i == p) {
      continue;
    }
    dfs(i, v);
    for (int j = 1; j <= k; j++) {
      d[v][j] += d[i][j - 1];
    }
  }
  ans += d[v][k];
  long long t = 0;
  for (int i : g[v]) {
    if (i == p) {
      continue;
    }
    for (int j = 0; j + 2 <= k; j++) {
      t += 1LL * d[i][j] * (d[v][k - j - 1] - d[i][k - j - 2]);
    }
  }
  t /= 2;
  ans += t;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a;
    --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int i = 0; i < n; i++) {
    d[i][0] = 1;
  }
  dfs(0, -1);
  cout << ans;
  return 0;
}

```