#include <bits/stdc++.h>
using namespace std;
int n, k;
vector<int> G[50001];
vector<int> childs;
bool visited[50001];
int size[50001];
int sum[501];
int level[50001];
int root;
int result;
void subtree_size(int u) {
  visited[u] = 1;
  size[u] = 1;
  for (int i = 0; i < (int)(((int)(G[u]).size())); i++) {
    int v = G[u][i];
    if (!visited[v]) {
      subtree_size(v);
      size[u] += size[v];
    }
  }
  visited[u] = 0;
}
void dfs(int u) {
  visited[u] = 1;
  childs.push_back(u);
  for (int i = 0; i < (int)(((int)(G[u]).size())); i++) {
    int v = G[u][i];
    if (!visited[v]) {
      level[v] = level[u] + 1;
      dfs(v);
    }
  }
  visited[u] = 0;
}
int findcentroid(int u) {
  visited[u] = 1;
  bool centroid = ((size[root] - size[u]) <= size[root] / 2);
  for (int i = 0; i < (int)(((int)(G[u]).size())); i++) {
    int v = G[u][i];
    if (!visited[v]) {
      centroid &= (size[v] <= size[root] / 2);
    }
  }
  if (centroid) {
    visited[u] = 0;
    return u;
  }
  for (int i = 0; i < (int)(((int)(G[u]).size())); i++) {
    int v = G[u][i];
    if (!visited[v]) {
      int c = findcentroid(v);
      if (c != -1) {
        visited[u] = 0;
        return c;
      }
    }
  }
  visited[u] = 0;
  return -1;
}
void find_k(int u) {
  subtree_size(u);
  root = u;
  u = findcentroid(u);
  for (int i = 0; i < (int)(k); i++) {
    sum[i] = 0;
  }
  sum[0] = 1;
  visited[u] = 1;
  for (int i = 0; i < (int)(((int)(G[u]).size())); i++) {
    int v = G[u][i];
    if (!visited[v]) {
      childs.clear();
      level[v] = 1;
      dfs(v);
      for (int j = 0; j < (int)(((int)(childs).size())); j++) {
        int w = childs[j];
        if (level[w] <= k) result += sum[k - level[w]];
      }
      for (int j = 0; j < (int)(((int)(childs).size())); j++) {
        int w = childs[j];
        if (level[w] <= k) sum[level[w]]++;
      }
    }
  }
  for (int i = 0; i < (int)(((int)(G[u]).size())); i++) {
    int v = G[u][i];
    if (!visited[v]) {
      find_k(v);
    }
  }
  visited[u] = 0;
}
int main() {
  int u, v;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    u--;
    v--;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  find_k(0);
  cout << result << endl;
}

### Prompt

Your challenge is to write 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;
int n, k;
vector<int> G[50001];
vector<int> childs;
bool visited[50001];
int size[50001];
int sum[501];
int level[50001];
int root;
int result;
void subtree_size(int u) {
  visited[u] = 1;
  size[u] = 1;
  for (int i = 0; i < (int)(((int)(G[u]).size())); i++) {
    int v = G[u][i];
    if (!visited[v]) {
      subtree_size(v);
      size[u] += size[v];
    }
  }
  visited[u] = 0;
}
void dfs(int u) {
  visited[u] = 1;
  childs.push_back(u);
  for (int i = 0; i < (int)(((int)(G[u]).size())); i++) {
    int v = G[u][i];
    if (!visited[v]) {
      level[v] = level[u] + 1;
      dfs(v);
    }
  }
  visited[u] = 0;
}
int findcentroid(int u) {
  visited[u] = 1;
  bool centroid = ((size[root] - size[u]) <= size[root] / 2);
  for (int i = 0; i < (int)(((int)(G[u]).size())); i++) {
    int v = G[u][i];
    if (!visited[v]) {
      centroid &= (size[v] <= size[root] / 2);
    }
  }
  if (centroid) {
    visited[u] = 0;
    return u;
  }
  for (int i = 0; i < (int)(((int)(G[u]).size())); i++) {
    int v = G[u][i];
    if (!visited[v]) {
      int c = findcentroid(v);
      if (c != -1) {
        visited[u] = 0;
        return c;
      }
    }
  }
  visited[u] = 0;
  return -1;
}
void find_k(int u) {
  subtree_size(u);
  root = u;
  u = findcentroid(u);
  for (int i = 0; i < (int)(k); i++) {
    sum[i] = 0;
  }
  sum[0] = 1;
  visited[u] = 1;
  for (int i = 0; i < (int)(((int)(G[u]).size())); i++) {
    int v = G[u][i];
    if (!visited[v]) {
      childs.clear();
      level[v] = 1;
      dfs(v);
      for (int j = 0; j < (int)(((int)(childs).size())); j++) {
        int w = childs[j];
        if (level[w] <= k) result += sum[k - level[w]];
      }
      for (int j = 0; j < (int)(((int)(childs).size())); j++) {
        int w = childs[j];
        if (level[w] <= k) sum[level[w]]++;
      }
    }
  }
  for (int i = 0; i < (int)(((int)(G[u]).size())); i++) {
    int v = G[u][i];
    if (!visited[v]) {
      find_k(v);
    }
  }
  visited[u] = 0;
}
int main() {
  int u, v;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    u--;
    v--;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  find_k(0);
  cout << result << endl;
}

```