#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:60777216")
using namespace std;
int n, k;
int a[1 << 20];
int d[1 << 20];
bitset<1 << 20> u;
bool good(int t) {
  u[0] = 1;
  for (int i = (0); i < (1 << 20); i++)
    if (d[i]) {
      int z = d[i];
      int j = 1;
      while (z) {
        j = min(j, z);
        z -= j;
        u = u | (u << (j * i));
        j *= 2;
      }
    }
  return u[t];
}
int main() {
  cin >> n >> k;
  for (int i = (0); i < (n); i++) scanf("%d", a + i), a[i]--;
  for (int i = (0); i < (n); i++)
    if (a[i] != -1) {
      int t = i;
      int num = 0;
      while (a[t] != -1) {
        int nt = a[t];
        a[t] = -1;
        t = nt;
        num++;
      }
      d[num]++;
    }
  int mn = k;
  int mx = 0;
  int r = k;
  for (int i = (0); i < (1 << 20); i++) {
    int z = min(i / 2 * d[i], r);
    mx += z * 2;
    r -= z;
  }
  for (int i = (0); i < (1 << 20); i++)
    if (i % 2 == 1 && r) {
      int z = min(d[i], r);
      r -= z;
      mx += z;
    }
  int tk = k;
  if (tk > n - k) tk = n - k;
  if (!good(tk)) mn++;
  cout << mn << ' ' << mx << endl;
  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>
#pragma comment(linker, "/STACK:60777216")
using namespace std;
int n, k;
int a[1 << 20];
int d[1 << 20];
bitset<1 << 20> u;
bool good(int t) {
  u[0] = 1;
  for (int i = (0); i < (1 << 20); i++)
    if (d[i]) {
      int z = d[i];
      int j = 1;
      while (z) {
        j = min(j, z);
        z -= j;
        u = u | (u << (j * i));
        j *= 2;
      }
    }
  return u[t];
}
int main() {
  cin >> n >> k;
  for (int i = (0); i < (n); i++) scanf("%d", a + i), a[i]--;
  for (int i = (0); i < (n); i++)
    if (a[i] != -1) {
      int t = i;
      int num = 0;
      while (a[t] != -1) {
        int nt = a[t];
        a[t] = -1;
        t = nt;
        num++;
      }
      d[num]++;
    }
  int mn = k;
  int mx = 0;
  int r = k;
  for (int i = (0); i < (1 << 20); i++) {
    int z = min(i / 2 * d[i], r);
    mx += z * 2;
    r -= z;
  }
  for (int i = (0); i < (1 << 20); i++)
    if (i % 2 == 1 && r) {
      int z = min(d[i], r);
      r -= z;
      mx += z;
    }
  int tk = k;
  if (tk > n - k) tk = n - k;
  if (!good(tk)) mn++;
  cout << mn << ' ' << mx << endl;
  return 0;
}

```