#include <bits/stdc++.h>
const long long MOD = 1e9 + 7;
const long long INF = 1e9 + 7;
using namespace std;
inline long long read();
vector<int> g[50010];
int n, k;
long long dp[50010][510];
long long ans = 0;
void dfs(int u, int fa) {
  dp[u][0] = 1;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == fa) continue;
    dfs(v, u);
    for (int i = 0; i < k; i++) {
      ans += dp[u][i] * dp[v][k - 1 - i];
      if (i) dp[u][i] += dp[v][i - 1];
    }
  }
}
int main() {
  n = read(), k = read();
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  cout << ans;
  return 0;
}
inline long long read() {
  long long p = 1, ans = 0;
  char ch = getchar();
  while (ch < 48 || ch > 57) {
    if (ch == '-') p = -1;
    ch = getchar();
  }
  while (ch >= 48 && ch <= 57) {
    ans = (ans << 3) + (ans << 1) + ch - 48;
    ch = getchar();
  }
  return ans * p;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
const long long MOD = 1e9 + 7;
const long long INF = 1e9 + 7;
using namespace std;
inline long long read();
vector<int> g[50010];
int n, k;
long long dp[50010][510];
long long ans = 0;
void dfs(int u, int fa) {
  dp[u][0] = 1;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == fa) continue;
    dfs(v, u);
    for (int i = 0; i < k; i++) {
      ans += dp[u][i] * dp[v][k - 1 - i];
      if (i) dp[u][i] += dp[v][i - 1];
    }
  }
}
int main() {
  n = read(), k = read();
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  cout << ans;
  return 0;
}
inline long long read() {
  long long p = 1, ans = 0;
  char ch = getchar();
  while (ch < 48 || ch > 57) {
    if (ch == '-') p = -1;
    ch = getchar();
  }
  while (ch >= 48 && ch <= 57) {
    ans = (ans << 3) + (ans << 1) + ch - 48;
    ch = getchar();
  }
  return ans * p;
}

```