#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 4;
int cnt[N];
int p[N];
int vis[N];
bitset<N> b;
int main() {
  b.set(0);
  int n, k;
  scanf("%d %d", &n, &k);
  int i;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
  }
  int odd = 0, pairs = 0;
  for (i = 1; i <= n; i++) {
    if (vis[i]) continue;
    int len = 0;
    int u = i;
    while (!vis[u]) {
      vis[u] = 1;
      len++;
      u = p[u];
    }
    cnt[len]++;
    odd += len % 2;
    pairs += len / 2;
  }
  for (i = 1; i <= n; i++) {
    if (cnt[i]) {
      int x = 1;
      while (cnt[i] >= x) {
        b |= (b << (i * x));
        cnt[i] -= x;
        x *= 2;
      }
      b |= (b << (cnt[i] * i));
    }
  }
  int mn, mx;
  if (b[k] == 1)
    mn = k;
  else
    mn = k + 1;
  if (pairs >= k)
    mx = 2 * k;
  else
    mx = 2 * pairs + min(k - pairs, odd);
  printf("%d %d\n", mn, mx);
  return 0;
}

### Prompt

Your challenge is to write 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;
const int N = 1e6 + 4;
int cnt[N];
int p[N];
int vis[N];
bitset<N> b;
int main() {
  b.set(0);
  int n, k;
  scanf("%d %d", &n, &k);
  int i;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
  }
  int odd = 0, pairs = 0;
  for (i = 1; i <= n; i++) {
    if (vis[i]) continue;
    int len = 0;
    int u = i;
    while (!vis[u]) {
      vis[u] = 1;
      len++;
      u = p[u];
    }
    cnt[len]++;
    odd += len % 2;
    pairs += len / 2;
  }
  for (i = 1; i <= n; i++) {
    if (cnt[i]) {
      int x = 1;
      while (cnt[i] >= x) {
        b |= (b << (i * x));
        cnt[i] -= x;
        x *= 2;
      }
      b |= (b << (cnt[i] * i));
    }
  }
  int mn, mx;
  if (b[k] == 1)
    mn = k;
  else
    mn = k + 1;
  if (pairs >= k)
    mx = 2 * k;
  else
    mx = 2 * pairs + min(k - pairs, odd);
  printf("%d %d\n", mn, mx);
  return 0;
}

```