#include <bits/stdc++.h>
using namespace std;
int n, k, SI, NN, ans;
int d[1000005];
int f[1000005];
vector<int> v[200005];
int se(int x, int o, int p) {
  if (!o) SI++;
  int su = 1;
  for (int i = 0; i < v[x].size(); i++)
    if (!f[v[x][i]] && v[x][i] != p) su += se(v[x][i], o, x);
  if (!NN && o == 1 && su >= SI / 2) NN = x;
  return su;
}
void sol(int x, int p, int high) {
  if (high == k) {
    ans++;
    return;
  }
  if (high > k) return;
  if (d[k - high]) {
    ans += d[k - high];
  }
  for (int i = 0; i < v[x].size(); i++)
    if (!f[v[x][i]] && v[x][i] != p) sol(v[x][i], x, high + 1);
}
void sol1(int x, int p, int high) {
  if (high >= k) return;
  d[high]++;
  for (int i = 0; i < v[x].size(); i++)
    if (!f[v[x][i]] && v[x][i] != p) sol1(v[x][i], x, high + 1);
}
void div(int x) {
  SI = NN = 0;
  se(x, 0, -1);
  se(x, 1, -1);
  int NNN = NN;
  f[NNN] = 1;
  for (int i = 0; i < v[NNN].size(); i++) {
    if (!f[v[NNN][i]]) sol(v[NNN][i], -1, 1), sol1(v[NNN][i], -1, 1);
  }
  for (int j = 0; j <= k; j++) d[j] = 0;
  for (int i = 0; i < v[NNN].size(); i++) {
    if (!f[v[NNN][i]]) div(v[NNN][i]);
  }
  for (int j = 0; j <= k; j++) d[j] = 0;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  div(1);
  cout << ans << endl;
}

### 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;
int n, k, SI, NN, ans;
int d[1000005];
int f[1000005];
vector<int> v[200005];
int se(int x, int o, int p) {
  if (!o) SI++;
  int su = 1;
  for (int i = 0; i < v[x].size(); i++)
    if (!f[v[x][i]] && v[x][i] != p) su += se(v[x][i], o, x);
  if (!NN && o == 1 && su >= SI / 2) NN = x;
  return su;
}
void sol(int x, int p, int high) {
  if (high == k) {
    ans++;
    return;
  }
  if (high > k) return;
  if (d[k - high]) {
    ans += d[k - high];
  }
  for (int i = 0; i < v[x].size(); i++)
    if (!f[v[x][i]] && v[x][i] != p) sol(v[x][i], x, high + 1);
}
void sol1(int x, int p, int high) {
  if (high >= k) return;
  d[high]++;
  for (int i = 0; i < v[x].size(); i++)
    if (!f[v[x][i]] && v[x][i] != p) sol1(v[x][i], x, high + 1);
}
void div(int x) {
  SI = NN = 0;
  se(x, 0, -1);
  se(x, 1, -1);
  int NNN = NN;
  f[NNN] = 1;
  for (int i = 0; i < v[NNN].size(); i++) {
    if (!f[v[NNN][i]]) sol(v[NNN][i], -1, 1), sol1(v[NNN][i], -1, 1);
  }
  for (int j = 0; j <= k; j++) d[j] = 0;
  for (int i = 0; i < v[NNN].size(); i++) {
    if (!f[v[NNN][i]]) div(v[NNN][i]);
  }
  for (int j = 0; j <= k; j++) d[j] = 0;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  div(1);
  cout << ans << endl;
}

```