#include <bits/stdc++.h>
using namespace std;
long long z = 1e9 + 7;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (res * a) % z;
      b--;
    } else {
      a = (a * a) % z;
      b = b >> 1;
    }
  }
  return res;
}
long long n, k, ans = 0, p1;
long long dist[50005][505] = {0};
vector<long long> v[50001];
void dfs(long long u, long long p = -1) {
  dist[u][0] = 1;
  for (long long i = 0; i < v[u].size(); i++) {
    if (v[u][i] == p) continue;
    dfs(v[u][i], u);
    for (long long j = 1; j <= k + 1; j++) dist[u][j] += dist[v[u][i]][j - 1];
  }
  for (long long i = 0; i < v[u].size(); i++) {
    if (v[u][i] == p) continue;
    for (long long j = 1; j <= k - 1; j++)
      ans += (dist[u][j] - dist[v[u][i]][j - 1]) * dist[v[u][i]][k - j - 1];
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  while (t--) {
    cin >> n >> k;
    long long x, y;
    for (long long i = 1; i < n; i++) {
      cin >> x >> y;
      v[x].push_back(y);
      v[y].push_back(x);
    }
    dfs(1);
    ans /= 2;
    for (long long i = 1; i <= n; i++) ans += dist[i][k];
    cout << ans << "\n";
  }
}

### Prompt

Create a solution in CPP for 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;
long long z = 1e9 + 7;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (res * a) % z;
      b--;
    } else {
      a = (a * a) % z;
      b = b >> 1;
    }
  }
  return res;
}
long long n, k, ans = 0, p1;
long long dist[50005][505] = {0};
vector<long long> v[50001];
void dfs(long long u, long long p = -1) {
  dist[u][0] = 1;
  for (long long i = 0; i < v[u].size(); i++) {
    if (v[u][i] == p) continue;
    dfs(v[u][i], u);
    for (long long j = 1; j <= k + 1; j++) dist[u][j] += dist[v[u][i]][j - 1];
  }
  for (long long i = 0; i < v[u].size(); i++) {
    if (v[u][i] == p) continue;
    for (long long j = 1; j <= k - 1; j++)
      ans += (dist[u][j] - dist[v[u][i]][j - 1]) * dist[v[u][i]][k - j - 1];
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  while (t--) {
    cin >> n >> k;
    long long x, y;
    for (long long i = 1; i < n; i++) {
      cin >> x >> y;
      v[x].push_back(y);
      v[y].push_back(x);
    }
    dfs(1);
    ans /= 2;
    for (long long i = 1; i <= n; i++) ans += dist[i][k];
    cout << ans << "\n";
  }
}

```