#include <bits/stdc++.h>
using namespace std;
int n, p[1000010];
int k;
int chk[1000010];
int cyc[1000010];
bitset<1000010> D;
void solve() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", p + i);
  int C[2] = {};
  for (int i = 1; i <= n; i++) {
    if (chk[i]) continue;
    int t = i;
    chk[t] = 1;
    t = p[t];
    int len = 1;
    while (chk[t] == 0) {
      chk[t] = 1;
      t = p[t];
      ++len;
    }
    C[1] += len / 2;
    C[0] += len % 2;
    cyc[len]++;
  }
  int mx = 0;
  if (k <= C[1])
    mx = 2 * k;
  else if (k <= C[1] + C[0])
    mx = k + C[1];
  else
    mx = n;
  for (int i = 2; i <= n; i++) {
    int t = cyc[i];
    cyc[i] = 0;
    for (int j = 0; j < 20; j++) {
      if (t >= 1 << j) t -= 1 << j, cyc[i << j]++;
    }
    for (int j = 0; j < 20; j++)
      if (1 << j & t) cyc[i << j]++;
  }
  D[0] = 1;
  for (int i = 2; i <= n; i++) {
    while (cyc[i]--) {
      D = D | D << i;
    }
  }
  if (D[k])
    printf("%d ", k);
  else
    printf("%d ", k + 1);
  printf("%d", mx);
}
int main() {
  int Tc = 1;
  for (int tc = 1; tc <= Tc; tc++) {
    solve();
  }
  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;
int n, p[1000010];
int k;
int chk[1000010];
int cyc[1000010];
bitset<1000010> D;
void solve() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", p + i);
  int C[2] = {};
  for (int i = 1; i <= n; i++) {
    if (chk[i]) continue;
    int t = i;
    chk[t] = 1;
    t = p[t];
    int len = 1;
    while (chk[t] == 0) {
      chk[t] = 1;
      t = p[t];
      ++len;
    }
    C[1] += len / 2;
    C[0] += len % 2;
    cyc[len]++;
  }
  int mx = 0;
  if (k <= C[1])
    mx = 2 * k;
  else if (k <= C[1] + C[0])
    mx = k + C[1];
  else
    mx = n;
  for (int i = 2; i <= n; i++) {
    int t = cyc[i];
    cyc[i] = 0;
    for (int j = 0; j < 20; j++) {
      if (t >= 1 << j) t -= 1 << j, cyc[i << j]++;
    }
    for (int j = 0; j < 20; j++)
      if (1 << j & t) cyc[i << j]++;
  }
  D[0] = 1;
  for (int i = 2; i <= n; i++) {
    while (cyc[i]--) {
      D = D | D << i;
    }
  }
  if (D[k])
    printf("%d ", k);
  else
    printf("%d ", k + 1);
  printf("%d", mx);
}
int main() {
  int Tc = 1;
  for (int tc = 1; tc <= Tc; tc++) {
    solve();
  }
  return 0;
}

```