#include <bits/stdc++.h>
using namespace std;
template <class _Tp>
void ckmax(_Tp &a, _Tp b) {
  if (a < b) a = b;
}
template <class _Tp>
void ckmin(_Tp &a, _Tp b) {
  if (a > b) a = b;
}
template <class _Tp>
_Tp gcd(_Tp a, _Tp b) {
  return (b == 0) ? (a) : (gcd(b, a % b));
}
int read() {
  char ch = getchar();
  bool f = 1;
  int x = 0;
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') f = 0, ch = getchar();
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch & 15);
    ch = getchar();
  }
  return f ? x : -x;
}
const int inf = 1000000000;
const long long Inf = 1000000000000000000ll;
int n, k, col[1000050], cnt, to[1000050];
int p[1000050], ans, tot, a[1000050];
bool vis[1000050];
void bfs(int st) {
  queue<int> q;
  q.push(st);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    col[u] = st;
    cnt++;
    if (!col[to[u]]) q.push(to[u]);
  }
}
int main() {
  n = read();
  k = read();
  for (int i = 1; i <= n; ++i) to[i] = read();
  for (int i = 1; i <= n; ++i)
    if (!col[i]) {
      cnt = 0;
      bfs(i);
      p[cnt]++;
      ans += (cnt >> 1);
    }
  for (int i = 1; i <= n; ++i) {
    if (!p[i]) continue;
    for (int j = 1; j <= p[i]; j <<= 1) a[++tot] = j * i, p[i] -= j;
    if (p[i]) a[++tot] = p[i] * i;
  }
  vis[0] = 1;
  for (int i = 1; i <= tot; ++i)
    for (int j = k - a[i]; j >= 0; --j)
      if (vis[j]) vis[j + a[i]] = 1;
  if (vis[k])
    printf("%d ", k);
  else
    printf("%d ", k + 1);
  if (ans >= k)
    printf("%d\n", k * 2);
  else
    printf("%d\n", ans * 2 + min(k - ans, n - ans * 2));
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

It's Christmas time! PolandBall and his friends will be giving themselves gifts. There are n Balls overall. Each Ball has someone for whom he should bring a present according to some permutation p, pi ≠ i for all i.

Unfortunately, Balls are quite clumsy. We know earlier that exactly k of them will forget to bring their gift. A Ball number i will get his present if the following two constraints will hold: 

  1. Ball number i will bring the present he should give. 
  2. Ball x such that px = i will bring his present. 



What is minimum and maximum possible number of kids who will not get their present if exactly k Balls will forget theirs?

Input

The first line of input contains two integers n and k (2 ≤ n ≤ 106, 0 ≤ k ≤ n), representing the number of Balls and the number of Balls who will forget to bring their presents. 

The second line contains the permutation p of integers from 1 to n, where pi is the index of Ball who should get a gift from the i-th Ball. For all i, pi ≠ i holds.

Output

You should output two values — minimum and maximum possible number of Balls who will not get their presents, in that order.

Examples

Input

5 2
3 4 1 5 2


Output

2 4

Input

10 1
2 3 4 5 6 7 8 9 10 1


Output

2 2

Note

In the first sample, if the third and the first balls will forget to bring their presents, they will be th only balls not getting a present. Thus the minimum answer is 2. However, if the first ans the second balls will forget to bring their presents, then only the fifth ball will get a present. So, the maximum answer is 4.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class _Tp>
void ckmax(_Tp &a, _Tp b) {
  if (a < b) a = b;
}
template <class _Tp>
void ckmin(_Tp &a, _Tp b) {
  if (a > b) a = b;
}
template <class _Tp>
_Tp gcd(_Tp a, _Tp b) {
  return (b == 0) ? (a) : (gcd(b, a % b));
}
int read() {
  char ch = getchar();
  bool f = 1;
  int x = 0;
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') f = 0, ch = getchar();
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch & 15);
    ch = getchar();
  }
  return f ? x : -x;
}
const int inf = 1000000000;
const long long Inf = 1000000000000000000ll;
int n, k, col[1000050], cnt, to[1000050];
int p[1000050], ans, tot, a[1000050];
bool vis[1000050];
void bfs(int st) {
  queue<int> q;
  q.push(st);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    col[u] = st;
    cnt++;
    if (!col[to[u]]) q.push(to[u]);
  }
}
int main() {
  n = read();
  k = read();
  for (int i = 1; i <= n; ++i) to[i] = read();
  for (int i = 1; i <= n; ++i)
    if (!col[i]) {
      cnt = 0;
      bfs(i);
      p[cnt]++;
      ans += (cnt >> 1);
    }
  for (int i = 1; i <= n; ++i) {
    if (!p[i]) continue;
    for (int j = 1; j <= p[i]; j <<= 1) a[++tot] = j * i, p[i] -= j;
    if (p[i]) a[++tot] = p[i] * i;
  }
  vis[0] = 1;
  for (int i = 1; i <= tot; ++i)
    for (int j = k - a[i]; j >= 0; --j)
      if (vis[j]) vis[j + a[i]] = 1;
  if (vis[k])
    printf("%d ", k);
  else
    printf("%d ", k + 1);
  if (ans >= k)
    printf("%d\n", k * 2);
  else
    printf("%d\n", ans * 2 + min(k - ans, n - ans * 2));
  return 0;
}

```