#include <bits/stdc++.h>
using namespace std;
const int N = 1000 * 1000 + 10, SQ = 177;
int num[N], p[N];
bitset<N> d, mark;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, k, a1(0), a2(0), mx(0), mn(0);
  scanf("%d %d", &n, &k);
  mn = k;
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
    p[i]--;
  }
  for (int i = 0; i < n; i++) {
    if (!mark[i]) {
      int pt(p[i]), cnt(0);
      while (pt != i) {
        mark[pt] = 1;
        pt = p[pt];
        cnt++;
      }
      mark[pt] = 1, cnt++;
      num[cnt]++;
      a1 += cnt / 2, a2 += cnt / 2;
      a2 += cnt % 2;
    }
  }
  if (k <= a1)
    mx = k * 2;
  else if (k <= a2)
    mx = a1 * 2 + (k - a1);
  else
    mx = n;
  d[0] = 1;
  for (int i = SQ; i <= n; i++) {
    for (int j = 0; j < num[i]; j++) d |= (d << i);
  }
  for (int i = 2; i < SQ; i++) {
    if (num[i]) {
      for (int j = n; j >= 0; j--) {
        if (d[j]) {
          int t = num[i], pt(j + i);
          while (pt <= n && !d[pt] && t) {
            d[pt] = 1;
            t--;
            pt += i;
          }
        }
      }
    }
  }
  if (!d[k]) mn++;
  cout << mn << " " << mx << endl;
  return 0;
}

### Prompt

Please formulate 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;
const int N = 1000 * 1000 + 10, SQ = 177;
int num[N], p[N];
bitset<N> d, mark;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, k, a1(0), a2(0), mx(0), mn(0);
  scanf("%d %d", &n, &k);
  mn = k;
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
    p[i]--;
  }
  for (int i = 0; i < n; i++) {
    if (!mark[i]) {
      int pt(p[i]), cnt(0);
      while (pt != i) {
        mark[pt] = 1;
        pt = p[pt];
        cnt++;
      }
      mark[pt] = 1, cnt++;
      num[cnt]++;
      a1 += cnt / 2, a2 += cnt / 2;
      a2 += cnt % 2;
    }
  }
  if (k <= a1)
    mx = k * 2;
  else if (k <= a2)
    mx = a1 * 2 + (k - a1);
  else
    mx = n;
  d[0] = 1;
  for (int i = SQ; i <= n; i++) {
    for (int j = 0; j < num[i]; j++) d |= (d << i);
  }
  for (int i = 2; i < SQ; i++) {
    if (num[i]) {
      for (int j = n; j >= 0; j--) {
        if (d[j]) {
          int t = num[i], pt(j + i);
          while (pt <= n && !d[pt] && t) {
            d[pt] = 1;
            t--;
            pt += i;
          }
        }
      }
    }
  }
  if (!d[k]) mn++;
  cout << mn << " " << mx << endl;
  return 0;
}

```