#include <bits/stdc++.h>
using namespace std;
int p[1000000 + 5], cc[1000000 + 5], cnt[3];
bool mark[1000000 + 5];
bitset<1000000 + 5> dp;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 1; i <= n; i++)
    if (mark[i] == 0) {
      int now = i, c = 0;
      while (mark[now] == 0) {
        c++;
        mark[now] = 1;
        now = p[now];
      }
      cnt[0] += c / 2;
      cnt[2] += c / 2;
      cnt[1] += c % 2;
      cc[c]++;
    }
  int mx = 0, K = k;
  for (int i = 2; i >= 0; i--) mx += min(K, cnt[i]) * i, K -= min(K, cnt[i]);
  dp[0] = 1;
  for (int i = 1; i <= n; i++)
    if (cc[i]) {
      for (int j = 0; j <= 25; j++) {
        if (cc[i] >= (1 << j))
          dp = dp | (dp << (i * (1 << j))), cc[i] -= 1 << j;
        else {
          dp = dp | (dp << (i * cc[i]));
          break;
        }
      }
    }
  printf("%d %d\n", dp[k] == 1 ? k : k + 1, mx);
  return 0;
}

### Prompt

In cpp, your task is to solve 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 p[1000000 + 5], cc[1000000 + 5], cnt[3];
bool mark[1000000 + 5];
bitset<1000000 + 5> dp;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 1; i <= n; i++)
    if (mark[i] == 0) {
      int now = i, c = 0;
      while (mark[now] == 0) {
        c++;
        mark[now] = 1;
        now = p[now];
      }
      cnt[0] += c / 2;
      cnt[2] += c / 2;
      cnt[1] += c % 2;
      cc[c]++;
    }
  int mx = 0, K = k;
  for (int i = 2; i >= 0; i--) mx += min(K, cnt[i]) * i, K -= min(K, cnt[i]);
  dp[0] = 1;
  for (int i = 1; i <= n; i++)
    if (cc[i]) {
      for (int j = 0; j <= 25; j++) {
        if (cc[i] >= (1 << j))
          dp = dp | (dp << (i * (1 << j))), cc[i] -= 1 << j;
        else {
          dp = dp | (dp << (i * cc[i]));
          break;
        }
      }
    }
  printf("%d %d\n", dp[k] == 1 ? k : k + 1, mx);
  return 0;
}

```