#include <bits/stdc++.h>
using namespace std;
int n, k;
namespace PDC {
vector<int> to[50005];
bool vis[50005];
int size[50005];
long long Ans = 0;
void add(int u, int v) {
  to[u].push_back(v);
  to[v].push_back(u);
}
int Min = 1e9, Root = 0, All = 0;
void find_root(int x, int fa) {
  int Max = 0;
  size[x] = 1;
  for (auto E : to[x])
    if (!vis[E] and E ^ fa) {
      find_root(E, x);
      size[x] += size[E];
      Max = max(Max, size[E]);
    }
  Max = max(Max, All - size[x]);
  if (Max < Min) Min = Max, Root = x;
}
int dis[50005], dis_cnt = 0;
void get_dist(int x, int fa, int f) {
  dis[++dis_cnt] = f;
  for (auto E : to[x])
    if (!vis[E] and E ^ fa) {
      get_dist(E, x, f + 1);
    }
}
void solve(int x, int con) {
  dis_cnt = 0;
  get_dist(x, x, con);
  sort(dis + 1, dis + dis_cnt + 1);
  int L = 1, R = dis_cnt;
  long long COL = 1, COR = 1;
  while (L <= R) {
    while (dis[L + 1] == dis[L] and L < dis_cnt) L++, COL++;
    while (dis[R - 1] == dis[R] and R >= 2) R--, COR++;
    if (dis[L] + dis[R] == k)
      if (dis[L] == dis[R]) Ans -= (con ? -1ll : 1ll) * (COL * COL + COL) / 2;
    if (dis[L] + dis[R] == k) Ans += (con ? -1ll : 1ll) * COL * COR;
    if (dis[L] + dis[R] < k)
      L++, COL = 1;
    else
      R--, COR = 1;
  }
}
void Divice(int x) {
  vis[x] = true;
  solve(x, 0);
  for (auto E : to[x])
    if (!vis[E]) {
      Min = 1e9, All = size[E];
      solve(E, 1);
      find_root(E, E);
      Divice(Root);
    }
}
}  // namespace PDC
int main() {
  cin >> n >> k;
  PDC::All = n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    PDC::add(u, v);
  }
  PDC::find_root(1, 1);
  PDC::Divice(PDC::Root);
  cout << PDC::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;
int n, k;
namespace PDC {
vector<int> to[50005];
bool vis[50005];
int size[50005];
long long Ans = 0;
void add(int u, int v) {
  to[u].push_back(v);
  to[v].push_back(u);
}
int Min = 1e9, Root = 0, All = 0;
void find_root(int x, int fa) {
  int Max = 0;
  size[x] = 1;
  for (auto E : to[x])
    if (!vis[E] and E ^ fa) {
      find_root(E, x);
      size[x] += size[E];
      Max = max(Max, size[E]);
    }
  Max = max(Max, All - size[x]);
  if (Max < Min) Min = Max, Root = x;
}
int dis[50005], dis_cnt = 0;
void get_dist(int x, int fa, int f) {
  dis[++dis_cnt] = f;
  for (auto E : to[x])
    if (!vis[E] and E ^ fa) {
      get_dist(E, x, f + 1);
    }
}
void solve(int x, int con) {
  dis_cnt = 0;
  get_dist(x, x, con);
  sort(dis + 1, dis + dis_cnt + 1);
  int L = 1, R = dis_cnt;
  long long COL = 1, COR = 1;
  while (L <= R) {
    while (dis[L + 1] == dis[L] and L < dis_cnt) L++, COL++;
    while (dis[R - 1] == dis[R] and R >= 2) R--, COR++;
    if (dis[L] + dis[R] == k)
      if (dis[L] == dis[R]) Ans -= (con ? -1ll : 1ll) * (COL * COL + COL) / 2;
    if (dis[L] + dis[R] == k) Ans += (con ? -1ll : 1ll) * COL * COR;
    if (dis[L] + dis[R] < k)
      L++, COL = 1;
    else
      R--, COR = 1;
  }
}
void Divice(int x) {
  vis[x] = true;
  solve(x, 0);
  for (auto E : to[x])
    if (!vis[E]) {
      Min = 1e9, All = size[E];
      solve(E, 1);
      find_root(E, E);
      Divice(Root);
    }
}
}  // namespace PDC
int main() {
  cin >> n >> k;
  PDC::All = n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    PDC::add(u, v);
  }
  PDC::find_root(1, 1);
  PDC::Divice(PDC::Root);
  cout << PDC::Ans;
  return 0;
}

```