#include <bits/stdc++.h>
using namespace std;
const int N = 50020;
struct Edge {
  int ver, next;
} e[N << 1];
int n, k, t, Head[N], d[N], flag[N], Max[N], sz[N], buc[N], rt, tot;
long long Ans;
inline void Insert(int x, int y) { e[++t] = (Edge){y, Head[x]}, Head[x] = t; }
inline void Findroot(int x, int fa) {
  sz[x] = 1, Max[x] = 0;
  for (int i = Head[x], y; i; i = e[i].next)
    if ((y = e[i].ver) != fa && !flag[y])
      Findroot(y, x), sz[x] += sz[y], Max[x] = max(Max[x], sz[y]);
  Max[x] = max(Max[x], tot - sz[x]);
  if (Max[rt] > Max[x]) rt = x;
}
inline void Calc(int x, int fa, int dis) {
  dis > k ?: Ans += buc[k - dis];
  for (int i = Head[x], y; i; i = e[i].next)
    if ((y = e[i].ver) != fa && !flag[y]) Calc(y, x, dis + 1);
}
inline void Modify(int x, int fa, int v) {
  d[x] = d[fa] + 1, buc[d[x]] += v;
  for (int i = Head[x], y; i; i = e[i].next)
    if ((y = e[i].ver) != fa && !flag[y]) Modify(y, x, v);
}
inline void Divide(int x) {
  flag[x] = true, d[x] = 0, buc[0] = 1;
  for (int i = Head[x], y; i; i = e[i].next)
    if (!flag[y = e[i].ver]) Calc(y, 0, 1), Modify(y, 0, 1);
  for (int i = Head[x], y; i; i = e[i].next)
    if (!flag[y = e[i].ver]) Modify(y, 0, -1);
  for (int i = Head[x], y; i; i = e[i].next)
    if (!flag[y = e[i].ver]) tot = sz[y], rt = 0, Findroot(y, 0), Divide(rt);
}
int main(void) {
  scanf("%d%d", &n, &k), Max[rt = 0] = 1e9, tot = n;
  for (int i = 1, u, v; i < n; i++)
    scanf("%d%d", &u, &v), Insert(u, v), Insert(v, u);
  return Findroot(1, 0), Divide(rt), printf("%lld\n", Ans), 0;
}

### Prompt

Your challenge is to write 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 int N = 50020;
struct Edge {
  int ver, next;
} e[N << 1];
int n, k, t, Head[N], d[N], flag[N], Max[N], sz[N], buc[N], rt, tot;
long long Ans;
inline void Insert(int x, int y) { e[++t] = (Edge){y, Head[x]}, Head[x] = t; }
inline void Findroot(int x, int fa) {
  sz[x] = 1, Max[x] = 0;
  for (int i = Head[x], y; i; i = e[i].next)
    if ((y = e[i].ver) != fa && !flag[y])
      Findroot(y, x), sz[x] += sz[y], Max[x] = max(Max[x], sz[y]);
  Max[x] = max(Max[x], tot - sz[x]);
  if (Max[rt] > Max[x]) rt = x;
}
inline void Calc(int x, int fa, int dis) {
  dis > k ?: Ans += buc[k - dis];
  for (int i = Head[x], y; i; i = e[i].next)
    if ((y = e[i].ver) != fa && !flag[y]) Calc(y, x, dis + 1);
}
inline void Modify(int x, int fa, int v) {
  d[x] = d[fa] + 1, buc[d[x]] += v;
  for (int i = Head[x], y; i; i = e[i].next)
    if ((y = e[i].ver) != fa && !flag[y]) Modify(y, x, v);
}
inline void Divide(int x) {
  flag[x] = true, d[x] = 0, buc[0] = 1;
  for (int i = Head[x], y; i; i = e[i].next)
    if (!flag[y = e[i].ver]) Calc(y, 0, 1), Modify(y, 0, 1);
  for (int i = Head[x], y; i; i = e[i].next)
    if (!flag[y = e[i].ver]) Modify(y, 0, -1);
  for (int i = Head[x], y; i; i = e[i].next)
    if (!flag[y = e[i].ver]) tot = sz[y], rt = 0, Findroot(y, 0), Divide(rt);
}
int main(void) {
  scanf("%d%d", &n, &k), Max[rt = 0] = 1e9, tot = n;
  for (int i = 1, u, v; i < n; i++)
    scanf("%d%d", &u, &v), Insert(u, v), Insert(v, u);
  return Findroot(1, 0), Divide(rt), printf("%lld\n", Ans), 0;
}

```