#include <bits/stdc++.h>
using namespace std;
struct no {
  vector<int> to;
  int sum, p;
  bool saw;
  no() {
    sum = 1;
    p = -1;
  }
} tree[50000];
int root = 0, n, kk;
void dfs(int u = root, int p = -1) {
  tree[u].p = p;
  for (int v : tree[u].to)
    if (v != p) {
      dfs(v, u);
      tree[u].sum += tree[v].sum;
    }
}
int centroid(int u) {
  if (tree[u].sum == 1) return u;
  while (true) {
    int v = -1;
    for (int k : tree[u].to)
      if (!tree[k].saw && (v == -1 || tree[k].sum > tree[v].sum)) v = k;
    if (tree[v].sum <= (tree[u].sum >> 1)) return u;
    int t = tree[u].sum;
    tree[u].sum -= tree[v].sum;
    tree[v].sum = t;
    u = v;
  }
}
long long int cntDist[501], accDist[501];
void dfsDist(int u, int p, int d = 1) {
  if (d > 500) return;
  cntDist[d] += 1;
  accDist[d] += 1;
  for (int v : tree[u].to)
    if (!tree[v].saw && v != p) dfsDist(v, u, d + 1);
}
long long int ans = 0;
int solve(int u = root, int p = -1) {
  u = centroid(u);
  memset(accDist, 0, sizeof accDist);
  accDist[0] = 1;
  for (int v : tree[u].to)
    if (!tree[v].saw) {
      memset(cntDist, 0, sizeof cntDist);
      dfsDist(v, u);
      for (int i = 1; i <= kk; ++i)
        ans += cntDist[i] * (accDist[kk - i] - cntDist[kk - i]);
    }
  tree[u].saw = true;
  tree[u].p = p;
  vector<int> to;
  if (p != -1) to.push_back(p);
  for (int v : tree[u].to)
    if (!tree[v].saw) to.push_back(solve(v, u));
  tree[u].to.assign(to.begin(), to.end());
  return u;
}
long long int go() {
  dfs();
  root = solve();
  return ans;
}
int main() {
  scanf("%d %d", &n, &kk);
  for (int i = 1; i < n; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    tree[a - 1].to.push_back(b - 1);
    tree[b - 1].to.push_back(a - 1);
  }
  printf("%lld\n", go());
  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;
struct no {
  vector<int> to;
  int sum, p;
  bool saw;
  no() {
    sum = 1;
    p = -1;
  }
} tree[50000];
int root = 0, n, kk;
void dfs(int u = root, int p = -1) {
  tree[u].p = p;
  for (int v : tree[u].to)
    if (v != p) {
      dfs(v, u);
      tree[u].sum += tree[v].sum;
    }
}
int centroid(int u) {
  if (tree[u].sum == 1) return u;
  while (true) {
    int v = -1;
    for (int k : tree[u].to)
      if (!tree[k].saw && (v == -1 || tree[k].sum > tree[v].sum)) v = k;
    if (tree[v].sum <= (tree[u].sum >> 1)) return u;
    int t = tree[u].sum;
    tree[u].sum -= tree[v].sum;
    tree[v].sum = t;
    u = v;
  }
}
long long int cntDist[501], accDist[501];
void dfsDist(int u, int p, int d = 1) {
  if (d > 500) return;
  cntDist[d] += 1;
  accDist[d] += 1;
  for (int v : tree[u].to)
    if (!tree[v].saw && v != p) dfsDist(v, u, d + 1);
}
long long int ans = 0;
int solve(int u = root, int p = -1) {
  u = centroid(u);
  memset(accDist, 0, sizeof accDist);
  accDist[0] = 1;
  for (int v : tree[u].to)
    if (!tree[v].saw) {
      memset(cntDist, 0, sizeof cntDist);
      dfsDist(v, u);
      for (int i = 1; i <= kk; ++i)
        ans += cntDist[i] * (accDist[kk - i] - cntDist[kk - i]);
    }
  tree[u].saw = true;
  tree[u].p = p;
  vector<int> to;
  if (p != -1) to.push_back(p);
  for (int v : tree[u].to)
    if (!tree[v].saw) to.push_back(solve(v, u));
  tree[u].to.assign(to.begin(), to.end());
  return u;
}
long long int go() {
  dfs();
  root = solve();
  return ans;
}
int main() {
  scanf("%d %d", &n, &kk);
  for (int i = 1; i < n; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    tree[a - 1].to.push_back(b - 1);
    tree[b - 1].to.push_back(a - 1);
  }
  printf("%lld\n", go());
  return 0;
}

```