#include <bits/stdc++.h>
using namespace std;
int main() {
  std::ios::sync_with_stdio(false);
  int n, k;
  scanf("%d%d", &(n), &(k));
  static int p[int(1e6 + 5)];
  for (int i = 1, __R = n; i <= __R; i++) scanf("%d", &(p[i]));
  static bool vis[int(1e6 + 5)];
  static int cnt[int(1e6 + 5)];
  int ev = 0, od = 0;
  for (int i = 1, __R = n; i <= __R; i++)
    if (!vis[i]) {
      int x = i;
      int len = 0;
      while (!vis[x]) {
        vis[x] = true;
        len++;
        x = p[x];
      }
      cnt[len]++;
      if (len & 1) od++;
      ev += (len >> 1);
    }
  bitset<int(1e6 + 5)> dp;
  dp[0] = 1;
  for (int i = 1, __R = n; i <= __R; i++)
    if (cnt[i]) {
      int j = 1;
      while (cnt[i] > 0) {
        j = min(j, cnt[i]);
        int x = j * i;
        dp |= (dp << x);
        cnt[i] -= j;
        j <<= 1;
      }
    }
  int m = k + (1 - dp[k]);
  int t = min(k, ev);
  int M = (t << 1) + min(k - t, od);
  printf("%d %d\n", m, M);
  return 0;
}

### Prompt

Please create a solution in Cpp 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 main() {
  std::ios::sync_with_stdio(false);
  int n, k;
  scanf("%d%d", &(n), &(k));
  static int p[int(1e6 + 5)];
  for (int i = 1, __R = n; i <= __R; i++) scanf("%d", &(p[i]));
  static bool vis[int(1e6 + 5)];
  static int cnt[int(1e6 + 5)];
  int ev = 0, od = 0;
  for (int i = 1, __R = n; i <= __R; i++)
    if (!vis[i]) {
      int x = i;
      int len = 0;
      while (!vis[x]) {
        vis[x] = true;
        len++;
        x = p[x];
      }
      cnt[len]++;
      if (len & 1) od++;
      ev += (len >> 1);
    }
  bitset<int(1e6 + 5)> dp;
  dp[0] = 1;
  for (int i = 1, __R = n; i <= __R; i++)
    if (cnt[i]) {
      int j = 1;
      while (cnt[i] > 0) {
        j = min(j, cnt[i]);
        int x = j * i;
        dp |= (dp << x);
        cnt[i] -= j;
        j <<= 1;
      }
    }
  int m = k + (1 - dp[k]);
  int t = min(k, ev);
  int M = (t << 1) + min(k - t, od);
  printf("%d %d\n", m, M);
  return 0;
}

```