#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int res = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    res = (res << 3) + (res << 1) + ch - '0';
    ch = getchar();
  }
  return res * f;
}
void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 | '0');
}
const int N = 50005;
const int inf = 505;
int n, k, fir[N], nxt[N << 1], go[N << 1], cnt, size[N], maxp[N];
int judge[N], sum, rt, rem[N], now[N], dis[N], ans;
bool vis[N];
inline void Add(int u, int v) {
  nxt[++cnt] = fir[u];
  fir[u] = cnt;
  go[cnt] = v;
}
void getrt(int u, int fa) {
  size[u] = 1;
  maxp[u] = 0;
  for (int e = fir[u]; e; e = nxt[e]) {
    int v = go[e];
    if (vis[v] || v == fa) continue;
    getrt(v, u);
    size[u] += size[v];
    maxp[u] = max(maxp[u], size[v]);
  }
  maxp[u] = max(maxp[u], sum - size[u]);
  if (maxp[u] < maxp[rt]) rt = u;
}
void getdis(int u, int fa) {
  if (dis[u] <= k) now[++now[0]] = dis[u];
  for (int e = fir[u]; e; e = nxt[e]) {
    int v = go[e];
    if (vis[v] || fa == v) continue;
    dis[v] = dis[u] + 1;
    getdis(v, u);
  }
}
void calc(int u) {
  for (int e = fir[u]; e; e = nxt[e]) {
    int v = go[e];
    if (vis[v]) continue;
    now[0] = 0;
    dis[v] = 1;
    getdis(v, 0);
    for (int i = 1, len = now[0]; i <= len; i++) ans += judge[k - now[i]];
    for (int i = 1, len = now[0]; i <= len; i++)
      rem[++rem[0]] = now[i], judge[now[i]]++;
  }
  for (int i = 1, len = rem[0]; i <= len; i++) judge[rem[i]] = 0;
}
void solve(int u) {
  vis[u] = true;
  rem[0] = 0;
  rem[++rem[0]] = 1;
  judge[0] = 1;
  calc(u);
  for (int e = fir[u]; e; e = nxt[e]) {
    int v = go[e];
    if (vis[v]) continue;
    maxp[rt = 0] = n;
    sum = size[v];
    getrt(v, 0);
    solve(rt);
  }
}
int main() {
  n = read();
  k = read();
  int u, v;
  for (int i = 1; i < n; i++) {
    u = read();
    v = read();
    Add(u, v);
    Add(v, u);
  }
  maxp[rt = 0] = sum = n;
  getrt(1, 0);
  solve(rt);
  write(ans);
  return 0;
}

### Prompt

Generate 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;
inline int read() {
  int res = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    res = (res << 3) + (res << 1) + ch - '0';
    ch = getchar();
  }
  return res * f;
}
void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 | '0');
}
const int N = 50005;
const int inf = 505;
int n, k, fir[N], nxt[N << 1], go[N << 1], cnt, size[N], maxp[N];
int judge[N], sum, rt, rem[N], now[N], dis[N], ans;
bool vis[N];
inline void Add(int u, int v) {
  nxt[++cnt] = fir[u];
  fir[u] = cnt;
  go[cnt] = v;
}
void getrt(int u, int fa) {
  size[u] = 1;
  maxp[u] = 0;
  for (int e = fir[u]; e; e = nxt[e]) {
    int v = go[e];
    if (vis[v] || v == fa) continue;
    getrt(v, u);
    size[u] += size[v];
    maxp[u] = max(maxp[u], size[v]);
  }
  maxp[u] = max(maxp[u], sum - size[u]);
  if (maxp[u] < maxp[rt]) rt = u;
}
void getdis(int u, int fa) {
  if (dis[u] <= k) now[++now[0]] = dis[u];
  for (int e = fir[u]; e; e = nxt[e]) {
    int v = go[e];
    if (vis[v] || fa == v) continue;
    dis[v] = dis[u] + 1;
    getdis(v, u);
  }
}
void calc(int u) {
  for (int e = fir[u]; e; e = nxt[e]) {
    int v = go[e];
    if (vis[v]) continue;
    now[0] = 0;
    dis[v] = 1;
    getdis(v, 0);
    for (int i = 1, len = now[0]; i <= len; i++) ans += judge[k - now[i]];
    for (int i = 1, len = now[0]; i <= len; i++)
      rem[++rem[0]] = now[i], judge[now[i]]++;
  }
  for (int i = 1, len = rem[0]; i <= len; i++) judge[rem[i]] = 0;
}
void solve(int u) {
  vis[u] = true;
  rem[0] = 0;
  rem[++rem[0]] = 1;
  judge[0] = 1;
  calc(u);
  for (int e = fir[u]; e; e = nxt[e]) {
    int v = go[e];
    if (vis[v]) continue;
    maxp[rt = 0] = n;
    sum = size[v];
    getrt(v, 0);
    solve(rt);
  }
}
int main() {
  n = read();
  k = read();
  int u, v;
  for (int i = 1; i < n; i++) {
    u = read();
    v = read();
    Add(u, v);
    Add(v, u);
  }
  maxp[rt = 0] = sum = n;
  getrt(1, 0);
  solve(rt);
  write(ans);
  return 0;
}

```