#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-11;
const int inf = 0x7FFFFFFF;
template <class T>
inline void checkmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void checkmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
template <class T>
void show(T a, int n) {
  for (int i = 0; i < n; ++i) cout << a[i] << ' ';
  cout << endl;
}
template <class T>
void show(T a, int r, int l) {
  for (int i = 0; i < r; ++i) show(a[i], l);
  cout << endl;
}
const int N = 150005;
int to[N], nxt[N];
int vis[N], head[N];
int n, k, ec;
long long ans;
int cc[50001][501];
void add(int u, int v) {
  to[ec] = v;
  nxt[ec] = head[u];
  head[u] = ec++;
}
void dfs(int now) {
  vis[now] = true;
  int v, e, *src = cc[now], *dst;
  src[0]++;
  for (e = head[now]; e != -1; e = nxt[e]) {
    v = to[e];
    if (!vis[v]) {
      dfs(v);
      dst = cc[v];
      for (int i = 0; i <= k - 1; i++)
        ans += ((long long)src[i]) * dst[k - 1 - i];
      for (int i = 0; i <= k - 1; i++) src[i + 1] += dst[i];
    }
  }
}
int main() {
  while (cin >> n >> k) {
    memset(head, -1, sizeof(head));
    for (int i = 1; i < n; i++) {
      int a, b;
      cin >> a >> b;
      add(a, b);
      add(b, a);
    }
    dfs(1);
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please formulate a cpp solution to 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;
const double pi = acos(-1.0);
const double eps = 1e-11;
const int inf = 0x7FFFFFFF;
template <class T>
inline void checkmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void checkmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
template <class T>
void show(T a, int n) {
  for (int i = 0; i < n; ++i) cout << a[i] << ' ';
  cout << endl;
}
template <class T>
void show(T a, int r, int l) {
  for (int i = 0; i < r; ++i) show(a[i], l);
  cout << endl;
}
const int N = 150005;
int to[N], nxt[N];
int vis[N], head[N];
int n, k, ec;
long long ans;
int cc[50001][501];
void add(int u, int v) {
  to[ec] = v;
  nxt[ec] = head[u];
  head[u] = ec++;
}
void dfs(int now) {
  vis[now] = true;
  int v, e, *src = cc[now], *dst;
  src[0]++;
  for (e = head[now]; e != -1; e = nxt[e]) {
    v = to[e];
    if (!vis[v]) {
      dfs(v);
      dst = cc[v];
      for (int i = 0; i <= k - 1; i++)
        ans += ((long long)src[i]) * dst[k - 1 - i];
      for (int i = 0; i <= k - 1; i++) src[i + 1] += dst[i];
    }
  }
}
int main() {
  while (cin >> n >> k) {
    memset(head, -1, sizeof(head));
    for (int i = 1; i < n; i++) {
      int a, b;
      cin >> a >> b;
      add(a, b);
      add(b, a);
    }
    dfs(1);
    cout << ans << endl;
  }
  return 0;
}

```