#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const int mod = 998244353;
int Mod(long long x) { return x >= mod ? x - mod : x; }
long long read() {
  long long x = 0;
  int f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + (ch ^ 48);
  return x * f;
}
void print(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
const int N = 50010, M = N;
int n, k;
;
struct edge {
  int nxt, to, w;
} e[M << 1];
int head[N], ecnt = 1;
void add(int u, int v, long long w = 0) {
  e[++ecnt].w = w, e[ecnt].to = v, e[ecnt].nxt = head[u];
  head[u] = ecnt;
}
void add_edge(int u, int v, long long w = 0) {
  add(u, v, w), add(v, u, w * 1);
};
long long ans;
int root, mx[N], S, siz[N], vis[N];
void get_root(int u, int fa) {
  siz[u] = 1, mx[u] = 0;
  for (int i = head[u], v; i; i = e[i].nxt)
    if (!vis[v = e[i].to] && v != fa)
      get_root(v, u), siz[u] += siz[v], mx[u] = max(mx[u], siz[v]);
  mx[u] = max(mx[u], S - siz[u]);
  if (mx[u] < mx[root]) root = u;
}
long long dis[N], res[N];
int maxx, tot;
void dfs(int u, int fa, int dep) {
  maxx = max(maxx, dep);
  res[dep]++;
  for (int i = head[u], v; i; i = e[i].nxt)
    if (!vis[v = e[i].to] && v != fa) dfs(v, u, dep + 1);
}
void calc(int u) {
  dis[0] = 1, tot = 0;
  for (int i = head[u], v; i; i = e[i].nxt)
    if (!vis[v = e[i].to]) {
      maxx = 0, dfs(v, u, 1);
      for (int j = 0; j <= maxx; j++) ans = ans + res[j] * dis[k - j];
      for (int j = 0; j <= maxx; j++) dis[j] += res[j], res[j] = 0;
      tot = max(tot, maxx);
    }
  for (int i = 0; i <= tot; i++) dis[i] = 0;
}
void solve(int u) {
  get_root(u, 0), vis[u] = 1;
  calc(u);
  for (int i = head[u], v; i; i = e[i].nxt)
    if (!vis[v = e[i].to]) {
      root = 0, S = siz[v], get_root(v, u);
      solve(root);
    }
}
int main() {
  n = read(), k = read();
  for (int i = 1; i < n; i++) add_edge(read(), read());
  mx[root = 0] = S = n;
  get_root(1, 0), solve(root);
  print(ans), putchar('\n');
}

### 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;
const double pi = acos(-1.0);
const int mod = 998244353;
int Mod(long long x) { return x >= mod ? x - mod : x; }
long long read() {
  long long x = 0;
  int f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + (ch ^ 48);
  return x * f;
}
void print(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
const int N = 50010, M = N;
int n, k;
;
struct edge {
  int nxt, to, w;
} e[M << 1];
int head[N], ecnt = 1;
void add(int u, int v, long long w = 0) {
  e[++ecnt].w = w, e[ecnt].to = v, e[ecnt].nxt = head[u];
  head[u] = ecnt;
}
void add_edge(int u, int v, long long w = 0) {
  add(u, v, w), add(v, u, w * 1);
};
long long ans;
int root, mx[N], S, siz[N], vis[N];
void get_root(int u, int fa) {
  siz[u] = 1, mx[u] = 0;
  for (int i = head[u], v; i; i = e[i].nxt)
    if (!vis[v = e[i].to] && v != fa)
      get_root(v, u), siz[u] += siz[v], mx[u] = max(mx[u], siz[v]);
  mx[u] = max(mx[u], S - siz[u]);
  if (mx[u] < mx[root]) root = u;
}
long long dis[N], res[N];
int maxx, tot;
void dfs(int u, int fa, int dep) {
  maxx = max(maxx, dep);
  res[dep]++;
  for (int i = head[u], v; i; i = e[i].nxt)
    if (!vis[v = e[i].to] && v != fa) dfs(v, u, dep + 1);
}
void calc(int u) {
  dis[0] = 1, tot = 0;
  for (int i = head[u], v; i; i = e[i].nxt)
    if (!vis[v = e[i].to]) {
      maxx = 0, dfs(v, u, 1);
      for (int j = 0; j <= maxx; j++) ans = ans + res[j] * dis[k - j];
      for (int j = 0; j <= maxx; j++) dis[j] += res[j], res[j] = 0;
      tot = max(tot, maxx);
    }
  for (int i = 0; i <= tot; i++) dis[i] = 0;
}
void solve(int u) {
  get_root(u, 0), vis[u] = 1;
  calc(u);
  for (int i = head[u], v; i; i = e[i].nxt)
    if (!vis[v = e[i].to]) {
      root = 0, S = siz[v], get_root(v, u);
      solve(root);
    }
}
int main() {
  n = read(), k = read();
  for (int i = 1; i < n; i++) add_edge(read(), read());
  mx[root = 0] = S = n;
  get_root(1, 0), solve(root);
  print(ans), putchar('\n');
}

```