#include <bits/stdc++.h>
using namespace std;
int cnt[1000005];
int n, m;
int p[1000005];
bool bz[1000005];
int ans1, ans2;
bitset<1000005> b;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 1; i <= n; i++) {
    if (bz[i]) continue;
    bz[i] = 1;
    int now = p[i];
    int tot = 1;
    while (1) {
      if (now == i) break;
      bz[now] = 1;
      ++tot;
      now = p[now];
    }
    ++cnt[tot];
  }
  int last = m;
  for (int i = 1; i <= n; i++) {
    if (cnt[i] * (i / 2) >= last) {
      ans2 += last * 2;
      last = 0;
    } else {
      ans2 += cnt[i] * (i / 2) * 2;
      last -= cnt[i] * (i / 2);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (i % 2 == 0) continue;
    if (cnt[i] >= last) {
      ans2 += last;
      last = 0;
    } else {
      ans2 += cnt[i];
      last -= cnt[i];
    }
  }
  b[0] = 1;
  for (int i = 1; i <= n; i++) {
    if (!cnt[i]) continue;
    int now = 1;
    while (now <= cnt[i]) {
      b |= (b << (now * i));
      cnt[i] -= now;
      now *= 2;
    }
    b |= (b << (cnt[i] * i));
  }
  if (b[m])
    ans1 = m;
  else
    ans1 = m + 1;
  printf("%d %d", ans1, ans2);
  return 0;
}

### Prompt

Your challenge is to write a cpp solution 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;
int cnt[1000005];
int n, m;
int p[1000005];
bool bz[1000005];
int ans1, ans2;
bitset<1000005> b;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 1; i <= n; i++) {
    if (bz[i]) continue;
    bz[i] = 1;
    int now = p[i];
    int tot = 1;
    while (1) {
      if (now == i) break;
      bz[now] = 1;
      ++tot;
      now = p[now];
    }
    ++cnt[tot];
  }
  int last = m;
  for (int i = 1; i <= n; i++) {
    if (cnt[i] * (i / 2) >= last) {
      ans2 += last * 2;
      last = 0;
    } else {
      ans2 += cnt[i] * (i / 2) * 2;
      last -= cnt[i] * (i / 2);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (i % 2 == 0) continue;
    if (cnt[i] >= last) {
      ans2 += last;
      last = 0;
    } else {
      ans2 += cnt[i];
      last -= cnt[i];
    }
  }
  b[0] = 1;
  for (int i = 1; i <= n; i++) {
    if (!cnt[i]) continue;
    int now = 1;
    while (now <= cnt[i]) {
      b |= (b << (now * i));
      cnt[i] -= now;
      now *= 2;
    }
    b |= (b << (cnt[i] * i));
  }
  if (b[m])
    ans1 = m;
  else
    ans1 = m + 1;
  printf("%d %d", ans1, ans2);
  return 0;
}

```