#include <bits/stdc++.h>
using namespace std;
int n, k, a[((int)1001 * 1000)], maxi, mini;
vector<int> v;
int jaghool;
bitset<((int)1001 * 1000)> dp, mark;
int main() {
  scanf("%d%d", &n, &k);
  int k2 = k;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++)
    if (!mark[i]) {
      int x = a[i], p = 1;
      while (x != i) p++, mark[x] = 1, x = a[x];
      v.push_back(p);
      maxi += min(k2, p / 2) * 2, k2 -= min(k2, p / 2);
      if (p % 2) jaghool++;
    }
  maxi += min(k2, jaghool);
  dp[0] = 1;
  sort(v.begin(), v.end());
  v.push_back(-1);
  vector<int> v2;
  for (int i = 0, cnt = 1; i < v.size(); i++) {
    if (v[i] != v[i + 1]) {
      int x = v[i];
      while (cnt) {
        dp |= (dp << v[i]);
        for (int j = 0; j < ((int)20); j++) {
          if ((1 << j) <= cnt && cnt < (1 << (j + 1))) {
            cnt -= (1 << j);
            break;
          }
          dp |= (dp << (v[i] << j));
        }
      }
      cnt = 1;
    } else
      cnt++;
  }
  cout << k + 1 - dp[k] << " " << maxi << "\n";
  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;
int n, k, a[((int)1001 * 1000)], maxi, mini;
vector<int> v;
int jaghool;
bitset<((int)1001 * 1000)> dp, mark;
int main() {
  scanf("%d%d", &n, &k);
  int k2 = k;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++)
    if (!mark[i]) {
      int x = a[i], p = 1;
      while (x != i) p++, mark[x] = 1, x = a[x];
      v.push_back(p);
      maxi += min(k2, p / 2) * 2, k2 -= min(k2, p / 2);
      if (p % 2) jaghool++;
    }
  maxi += min(k2, jaghool);
  dp[0] = 1;
  sort(v.begin(), v.end());
  v.push_back(-1);
  vector<int> v2;
  for (int i = 0, cnt = 1; i < v.size(); i++) {
    if (v[i] != v[i + 1]) {
      int x = v[i];
      while (cnt) {
        dp |= (dp << v[i]);
        for (int j = 0; j < ((int)20); j++) {
          if ((1 << j) <= cnt && cnt < (1 << (j + 1))) {
            cnt -= (1 << j);
            break;
          }
          dp |= (dp << (v[i] << j));
        }
      }
      cnt = 1;
    } else
      cnt++;
  }
  cout << k + 1 - dp[k] << " " << maxi << "\n";
  return 0;
}

```