#include <bits/stdc++.h>
using namespace std;
static int visited[1000005], k, cyc[1000005];
bitset<1000005> dp;
int main() {
  static int loop, i, x, n, p[1000005], mi, ma, lo, op;
  scanf("%d %d", &n, &k);
  for (i = 0; i < n; i++) {
    scanf("%d", p + i);
    p[i]--;
  }
  lo = op = 0;
  for (i = 0; i < n; i++) {
    if (!visited[i]) {
      loop = 1;
      visited[i] = 1;
      x = p[i];
      while (!visited[x]) {
        loop++;
        visited[x] = 1;
        x = p[x];
      }
      cyc[loop]++;
      if (loop % 2) op++;
      lo += loop / 2;
    }
  }
  dp[0] = 1;
  for (int x = 1; x <= k; x++) {
    if (cyc[x] > 0) {
      i = 1;
      while (cyc[x] > 0) {
        i = min(i, cyc[x]);
        dp |= dp << (x * i);
        cyc[x] -= i;
        i *= 2;
      }
    }
  }
  if (dp[k] == 1)
    mi = k;
  else
    mi = k + 1;
  printf("%d %d\n", mi, 2 * min(lo, k) + min(k - min(lo, k), op));
  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;
static int visited[1000005], k, cyc[1000005];
bitset<1000005> dp;
int main() {
  static int loop, i, x, n, p[1000005], mi, ma, lo, op;
  scanf("%d %d", &n, &k);
  for (i = 0; i < n; i++) {
    scanf("%d", p + i);
    p[i]--;
  }
  lo = op = 0;
  for (i = 0; i < n; i++) {
    if (!visited[i]) {
      loop = 1;
      visited[i] = 1;
      x = p[i];
      while (!visited[x]) {
        loop++;
        visited[x] = 1;
        x = p[x];
      }
      cyc[loop]++;
      if (loop % 2) op++;
      lo += loop / 2;
    }
  }
  dp[0] = 1;
  for (int x = 1; x <= k; x++) {
    if (cyc[x] > 0) {
      i = 1;
      while (cyc[x] > 0) {
        i = min(i, cyc[x]);
        dp |= dp << (x * i);
        cyc[x] -= i;
        i *= 2;
      }
    }
  }
  if (dp[k] == 1)
    mi = k;
  else
    mi = k + 1;
  printf("%d %d\n", mi, 2 * min(lo, k) + min(k - min(lo, k), op));
  return 0;
}

```