#include <bits/stdc++.h>
using namespace std;
const int maxn = (int)5e4 + 13;
const long long INFL = 9223372036854775807;
const long long INF = 2147483647;
const long long MOD = (int)1e9 + 7, MOD2 = 998244353;
const long long base = 101, base2 = 37;
const long double pi = 3.1415926535897932;
vector<int> gr[maxn];
long long dp[maxn][600];
long long ans = 0;
int k;
void dfs(int v, int p) {
  dp[v][0] = 1;
  for (auto t : gr[v])
    if (t != p) dfs(t, v);
  for (int i = 1; i <= k; ++i) {
    for (auto t : gr[v])
      if (t != p) dp[v][i] += dp[t][i - 1];
  }
}
void res(int v, int p) {
  for (auto t : gr[v])
    if (t != p) res(t, v);
  long long h = 0;
  for (int i = 1; i <= k - 1; ++i) {
    for (auto t : gr[v]) {
      if (t != p) {
        h += dp[t][i - 1] * (dp[v][k - i] - dp[t][k - i - 1]);
      }
    }
  }
  ans += h / 2;
  ans += dp[v][k];
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  ;
  int n;
  cin >> n >> k;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    gr[u].push_back(v);
    gr[v].push_back(u);
  }
  if (k == 1) {
    cout << n - 1 << "\n";
    return 0;
  }
  dfs(1, 1);
  res(1, 1);
  cout << ans << "\n";
  return 0;
}

### 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>
using namespace std;
const int maxn = (int)5e4 + 13;
const long long INFL = 9223372036854775807;
const long long INF = 2147483647;
const long long MOD = (int)1e9 + 7, MOD2 = 998244353;
const long long base = 101, base2 = 37;
const long double pi = 3.1415926535897932;
vector<int> gr[maxn];
long long dp[maxn][600];
long long ans = 0;
int k;
void dfs(int v, int p) {
  dp[v][0] = 1;
  for (auto t : gr[v])
    if (t != p) dfs(t, v);
  for (int i = 1; i <= k; ++i) {
    for (auto t : gr[v])
      if (t != p) dp[v][i] += dp[t][i - 1];
  }
}
void res(int v, int p) {
  for (auto t : gr[v])
    if (t != p) res(t, v);
  long long h = 0;
  for (int i = 1; i <= k - 1; ++i) {
    for (auto t : gr[v]) {
      if (t != p) {
        h += dp[t][i - 1] * (dp[v][k - i] - dp[t][k - i - 1]);
      }
    }
  }
  ans += h / 2;
  ans += dp[v][k];
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  ;
  int n;
  cin >> n >> k;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    gr[u].push_back(v);
    gr[v].push_back(u);
  }
  if (k == 1) {
    cout << n - 1 << "\n";
    return 0;
  }
  dfs(1, 1);
  res(1, 1);
  cout << ans << "\n";
  return 0;
}

```