#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-8;
const double Pi = acos(-1);
bool inline equ(double a, double b) { return fabs(a - b) < EPS; }
int _R(int &x) { return scanf("%d", &x); }
int _R(long long &x) { return scanf("%" PRId64, &x); }
int _R(double &x) { return scanf("%lf", &x); }
int _R(char *s) { return scanf("%s", s); }
int R() { return 0; }
template <typename T, typename... U>
int R(T &x, U &...tail) {
  return _R(x) + R(tail...);
}
template <class T>
void _W(const T &x) {
  cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%" PRId64, x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T>
void _W(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _W(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
  _W(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  W(tail...);
}
const int N = 5e4 + 10;
const int K = 510;
int n, k;
vector<int> g[N];
long long cnt[N][K], ans;
void dfs(int u, int p) {
  cnt[u][0] = 1;
  for (int v : g[u])
    if (v != p) {
      dfs(v, u);
      for (int d = (0); d <= (k - 1); d++) {
        ans += cnt[u][k - d - 1] * cnt[v][d];
      }
      for (int d = (0); d <= (k - 1); d++) cnt[u][d + 1] += cnt[v][d];
    }
}
int main() {
  R(n, k);
  for (int i = 0; i < (n - 1); i++) {
    int u, v;
    R(u, v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, -1);
  W(ans);
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 double EPS = 1e-8;
const double Pi = acos(-1);
bool inline equ(double a, double b) { return fabs(a - b) < EPS; }
int _R(int &x) { return scanf("%d", &x); }
int _R(long long &x) { return scanf("%" PRId64, &x); }
int _R(double &x) { return scanf("%lf", &x); }
int _R(char *s) { return scanf("%s", s); }
int R() { return 0; }
template <typename T, typename... U>
int R(T &x, U &...tail) {
  return _R(x) + R(tail...);
}
template <class T>
void _W(const T &x) {
  cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%" PRId64, x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T>
void _W(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _W(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
  _W(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  W(tail...);
}
const int N = 5e4 + 10;
const int K = 510;
int n, k;
vector<int> g[N];
long long cnt[N][K], ans;
void dfs(int u, int p) {
  cnt[u][0] = 1;
  for (int v : g[u])
    if (v != p) {
      dfs(v, u);
      for (int d = (0); d <= (k - 1); d++) {
        ans += cnt[u][k - d - 1] * cnt[v][d];
      }
      for (int d = (0); d <= (k - 1); d++) cnt[u][d + 1] += cnt[v][d];
    }
}
int main() {
  R(n, k);
  for (int i = 0; i < (n - 1); i++) {
    int u, v;
    R(u, v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, -1);
  W(ans);
}

```