#include <bits/stdc++.h>
using namespace std;
const int sz = 1e6 + 100;
bitset<sz> b;
int p[sz], len[sz];
bool used[sz];
int main() {
  ios_base::sync_with_stdio(false);
  int n, k, sum = 0, cnt = 0, i;
  cin >> n >> k;
  for (i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
  }
  for (i = 0; i < n; i++)
    if (!used[i]) {
      int v = i, cursum = 0;
      while (!used[v]) {
        cursum++;
        used[v] = true;
        v = p[v];
      }
      len[cursum]++;
      sum += cursum >> 1;
      cnt += cursum & 1;
    }
  b[0] = 1;
  for (i = 1; i <= n; i++)
    if (len[i] != 0) {
      int j = 1;
      while (len[i] != 0) {
        j = min(len[i], j);
        int cur = j * i;
        b |= b << cur;
        len[i] -= j;
        j <<= 1;
      }
    }
  int ans = k;
  if (!b[k]) ans++;
  cout << ans << " ";
  ans = 2 * min(k, sum);
  k -= min(k, sum);
  ans += min(k, cnt);
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 sz = 1e6 + 100;
bitset<sz> b;
int p[sz], len[sz];
bool used[sz];
int main() {
  ios_base::sync_with_stdio(false);
  int n, k, sum = 0, cnt = 0, i;
  cin >> n >> k;
  for (i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
  }
  for (i = 0; i < n; i++)
    if (!used[i]) {
      int v = i, cursum = 0;
      while (!used[v]) {
        cursum++;
        used[v] = true;
        v = p[v];
      }
      len[cursum]++;
      sum += cursum >> 1;
      cnt += cursum & 1;
    }
  b[0] = 1;
  for (i = 1; i <= n; i++)
    if (len[i] != 0) {
      int j = 1;
      while (len[i] != 0) {
        j = min(len[i], j);
        int cur = j * i;
        b |= b << cur;
        len[i] -= j;
        j <<= 1;
      }
    }
  int ans = k;
  if (!b[k]) ans++;
  cout << ans << " ";
  ans = 2 * min(k, sum);
  k -= min(k, sum);
  ans += min(k, cnt);
  cout << ans << endl;
  return 0;
}

```