#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  while ((c < '0') || (c > '9')) c = getchar();
  while ((c >= '0') && (c <= '9')) {
    x = (x << 1) + (x << 3) + (c ^ '0');
    c = getchar();
  }
}
long long head[50005], nxt[50005 << 1], to[50005 << 1], cnt;
inline void add(long long u, long long v) {
  nxt[cnt] = head[u];
  head[u] = cnt;
  to[cnt++] = v;
  nxt[cnt] = head[v];
  head[v] = cnt;
  to[cnt++] = u;
}
long long n, k, dis[50005], son[50005], siz[50005], top, sum, root, ans;
bool visit[50005];
inline void get_root(long long pos, long long pre) {
  siz[pos] = 1;
  son[pos] = 0;
  for (register long long i = head[pos]; ~i; i = nxt[i]) {
    if (to[i] == pre || visit[to[i]]) continue;
    get_root(to[i], pos);
    siz[pos] += siz[to[i]];
    son[pos] = max(son[pos], siz[to[i]]);
  }
  son[pos] = max(son[pos], sum - siz[pos]);
  if (son[pos] < son[root]) root = pos;
}
inline void get_dis(long long pos, long long pre, long long w) {
  dis[++top] = w;
  for (register long long i = head[pos]; ~i; i = nxt[i]) {
    if (to[i] == pre || visit[to[i]]) continue;
    if (w < k) get_dis(to[i], pos, w + 1);
  }
}
inline long long calc(long long pos, long long w) {
  top = 0;
  get_dis(pos, pos, w);
  sort(dis + 1, dis + top + 1);
  long long ret = 0, l = 1, r = 1;
  for (register long long i = top; i >= 1; i--) {
    while ((dis[r] <= k - dis[i]) && (r <= top)) r++;
    while ((dis[l] < k - dis[i]) && (l <= top)) l++;
    if (i < r - 1)
      break;
    else if (i == r - 1) {
      l = r - l;
      ret += (l * (l - 1) >> 1ll);
      break;
    }
    ret += r - l;
  }
  return ret;
}
inline void solve(int pos) {
  visit[pos] = 1;
  ans += calc(pos, 0);
  for (register long long i = head[pos]; ~i; i = nxt[i]) {
    if (visit[to[i]]) continue;
    ans -= calc(to[i], 1);
    sum = siz[to[i]];
    root = 0;
    get_root(to[i], to[i]);
    solve(root);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  read(n), read(k);
  memset(head, -1, sizeof(head));
  for (register long long i = 1, ui, vi; i < n; i++) {
    read(ui), read(vi);
    add(ui, vi);
  }
  son[0] = sum = n;
  get_root(1, 1);
  solve(root);
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Your task is to create 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;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  while ((c < '0') || (c > '9')) c = getchar();
  while ((c >= '0') && (c <= '9')) {
    x = (x << 1) + (x << 3) + (c ^ '0');
    c = getchar();
  }
}
long long head[50005], nxt[50005 << 1], to[50005 << 1], cnt;
inline void add(long long u, long long v) {
  nxt[cnt] = head[u];
  head[u] = cnt;
  to[cnt++] = v;
  nxt[cnt] = head[v];
  head[v] = cnt;
  to[cnt++] = u;
}
long long n, k, dis[50005], son[50005], siz[50005], top, sum, root, ans;
bool visit[50005];
inline void get_root(long long pos, long long pre) {
  siz[pos] = 1;
  son[pos] = 0;
  for (register long long i = head[pos]; ~i; i = nxt[i]) {
    if (to[i] == pre || visit[to[i]]) continue;
    get_root(to[i], pos);
    siz[pos] += siz[to[i]];
    son[pos] = max(son[pos], siz[to[i]]);
  }
  son[pos] = max(son[pos], sum - siz[pos]);
  if (son[pos] < son[root]) root = pos;
}
inline void get_dis(long long pos, long long pre, long long w) {
  dis[++top] = w;
  for (register long long i = head[pos]; ~i; i = nxt[i]) {
    if (to[i] == pre || visit[to[i]]) continue;
    if (w < k) get_dis(to[i], pos, w + 1);
  }
}
inline long long calc(long long pos, long long w) {
  top = 0;
  get_dis(pos, pos, w);
  sort(dis + 1, dis + top + 1);
  long long ret = 0, l = 1, r = 1;
  for (register long long i = top; i >= 1; i--) {
    while ((dis[r] <= k - dis[i]) && (r <= top)) r++;
    while ((dis[l] < k - dis[i]) && (l <= top)) l++;
    if (i < r - 1)
      break;
    else if (i == r - 1) {
      l = r - l;
      ret += (l * (l - 1) >> 1ll);
      break;
    }
    ret += r - l;
  }
  return ret;
}
inline void solve(int pos) {
  visit[pos] = 1;
  ans += calc(pos, 0);
  for (register long long i = head[pos]; ~i; i = nxt[i]) {
    if (visit[to[i]]) continue;
    ans -= calc(to[i], 1);
    sum = siz[to[i]];
    root = 0;
    get_root(to[i], to[i]);
    solve(root);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  read(n), read(k);
  memset(head, -1, sizeof(head));
  for (register long long i = 1, ui, vi; i < n; i++) {
    read(ui), read(vi);
    add(ui, vi);
  }
  son[0] = sum = n;
  get_root(1, 1);
  solve(root);
  printf("%I64d\n", ans);
  return 0;
}

```