#include <bits/stdc++.h>
using namespace std;
const int NMAX = 50010;
const int INF = 0x7fffffff;
vector<pair<int, int>> g[NMAX];
int siz[NMAX];
int n, K;
int dis[NMAX];
int f[NMAX];
bool used[NMAX];
int allnode, rt;
long long ans;
int GET_DIS(int u, int fa, int d) {
  dis[++dis[0]] = d;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i].first;
    int w = g[u][i].second;
    if (v == fa || used[v]) continue;
    GET_DIS(v, u, d + w);
  }
  return 0;
}
int GET_ROOT(int u, int fa) {
  siz[u] = 1;
  f[u] = 0;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i].first;
    if (v == fa || used[v]) continue;
    GET_ROOT(v, u);
    siz[u] += siz[v];
    f[u] = max(f[u], siz[v]);
  }
  f[u] = max(f[u], allnode - siz[u]);
  if (f[u] < f[rt]) rt = u;
  return rt;
}
long long GET_ANS(int u, int d) {
  dis[0] = 0;
  GET_DIS(u, -1, d);
  sort(dis + 1, dis + 1 + dis[0]);
  int p1 = dis[0], p2 = dis[0];
  long long ret = 0;
  for (int i = 1; i <= dis[0]; i++) {
    while (p2 >= 1 && dis[p2] + dis[i] > K) p2 -= 1;
    if (p1 > p2) p1 = p2;
    while (p1 >= 1 && dis[p1] + dis[i] >= K) p1 -= 1;
    ret += p2 - p1;
  }
  return ret;
}
int solve(int u) {
  used[u] = true;
  ans += GET_ANS(u, 0);
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i].first;
    int w = g[u][i].second;
    if (used[v]) continue;
    ans -= GET_ANS(v, w);
    allnode = siz[v];
    rt = 0;
    solve(GET_ROOT(v, -1));
  }
  return 0;
}
int main() {
  f[0] = INF;
  scanf("%d%d", &n, &K);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(make_pair(v, 1));
    g[v].push_back(make_pair(u, 1));
  }
  rt = 0;
  allnode = n;
  solve(GET_ROOT(1, -1));
  printf("%I64d\n", ans / 2);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 NMAX = 50010;
const int INF = 0x7fffffff;
vector<pair<int, int>> g[NMAX];
int siz[NMAX];
int n, K;
int dis[NMAX];
int f[NMAX];
bool used[NMAX];
int allnode, rt;
long long ans;
int GET_DIS(int u, int fa, int d) {
  dis[++dis[0]] = d;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i].first;
    int w = g[u][i].second;
    if (v == fa || used[v]) continue;
    GET_DIS(v, u, d + w);
  }
  return 0;
}
int GET_ROOT(int u, int fa) {
  siz[u] = 1;
  f[u] = 0;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i].first;
    if (v == fa || used[v]) continue;
    GET_ROOT(v, u);
    siz[u] += siz[v];
    f[u] = max(f[u], siz[v]);
  }
  f[u] = max(f[u], allnode - siz[u]);
  if (f[u] < f[rt]) rt = u;
  return rt;
}
long long GET_ANS(int u, int d) {
  dis[0] = 0;
  GET_DIS(u, -1, d);
  sort(dis + 1, dis + 1 + dis[0]);
  int p1 = dis[0], p2 = dis[0];
  long long ret = 0;
  for (int i = 1; i <= dis[0]; i++) {
    while (p2 >= 1 && dis[p2] + dis[i] > K) p2 -= 1;
    if (p1 > p2) p1 = p2;
    while (p1 >= 1 && dis[p1] + dis[i] >= K) p1 -= 1;
    ret += p2 - p1;
  }
  return ret;
}
int solve(int u) {
  used[u] = true;
  ans += GET_ANS(u, 0);
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i].first;
    int w = g[u][i].second;
    if (used[v]) continue;
    ans -= GET_ANS(v, w);
    allnode = siz[v];
    rt = 0;
    solve(GET_ROOT(v, -1));
  }
  return 0;
}
int main() {
  f[0] = INF;
  scanf("%d%d", &n, &K);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(make_pair(v, 1));
    g[v].push_back(make_pair(u, 1));
  }
  rt = 0;
  allnode = n;
  solve(GET_ROOT(1, -1));
  printf("%I64d\n", ans / 2);
  return 0;
}

```