#include <bits/stdc++.h>
using namespace std;
ifstream fin("input.txt");
ofstream fout("output.txt");
vector<int> g[50505];
long long dp[50505][555];
int n, k;
bool used[50505];
long long res;
void dfs(int ver, int pref) {
  used[ver] = true;
  for (int i = 0; i < g[ver].size(); i++) {
    int to = g[ver][i];
    if (pref == to) continue;
    dfs(to, ver);
    for (int j = 1; j <= k; j++) dp[ver][j] += dp[to][j - 1];
  }
}
void dfs_res(int ver, int pref) {
  res += dp[ver][k] * 2;
  for (int i = 0; i < g[ver].size(); i++) {
    int to = g[ver][i];
    if (pref == to) continue;
    dfs_res(to, ver);
    bool flag = true;
    for (int j = 1; j <= k - 1; j++)
      res += dp[to][k - j - 1] * (dp[ver][j] - dp[to][j - 1]);
  }
}
int main() {
  cin >> n >> k;
  memset(used, false, sizeof used);
  for (int i = 1; i <= n; i++) dp[i][0] = 1;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, -1);
  dfs_res(1, -1);
  cout << res / 2 << 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;
ifstream fin("input.txt");
ofstream fout("output.txt");
vector<int> g[50505];
long long dp[50505][555];
int n, k;
bool used[50505];
long long res;
void dfs(int ver, int pref) {
  used[ver] = true;
  for (int i = 0; i < g[ver].size(); i++) {
    int to = g[ver][i];
    if (pref == to) continue;
    dfs(to, ver);
    for (int j = 1; j <= k; j++) dp[ver][j] += dp[to][j - 1];
  }
}
void dfs_res(int ver, int pref) {
  res += dp[ver][k] * 2;
  for (int i = 0; i < g[ver].size(); i++) {
    int to = g[ver][i];
    if (pref == to) continue;
    dfs_res(to, ver);
    bool flag = true;
    for (int j = 1; j <= k - 1; j++)
      res += dp[to][k - j - 1] * (dp[ver][j] - dp[to][j - 1]);
  }
}
int main() {
  cin >> n >> k;
  memset(used, false, sizeof used);
  for (int i = 1; i <= n; i++) dp[i][0] = 1;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, -1);
  dfs_res(1, -1);
  cout << res / 2 << endl;
  return 0;
}

```