#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1E6 + 10;
int a[MAXN];
int cnt[MAXN];
bitset<MAXN> f;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  int ans2 = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] < 0) continue;
    int len = 0;
    for (int u = i; a[u] > 0; ++len) u = -(a[u] = -a[u]);
    ++cnt[len];
    ans2 += len >> 1;
  }
  ans2 = min(n, k + min(k, ans2));
  f.set(0);
  for (int i = 1; i <= n; ++i) {
    if (cnt[i] == 0) continue;
    for (int t = cnt[i]; t; t >>= 1) f |= f << (t + 1 >> 1) * i;
  }
  printf("%d %d\n", k + !f[k], ans2);
  return 0;
}

### Prompt

Create a solution in CPP for 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 MAXN = 1E6 + 10;
int a[MAXN];
int cnt[MAXN];
bitset<MAXN> f;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  int ans2 = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] < 0) continue;
    int len = 0;
    for (int u = i; a[u] > 0; ++len) u = -(a[u] = -a[u]);
    ++cnt[len];
    ans2 += len >> 1;
  }
  ans2 = min(n, k + min(k, ans2));
  f.set(0);
  for (int i = 1; i <= n; ++i) {
    if (cnt[i] == 0) continue;
    for (int t = cnt[i]; t; t >>= 1) f |= f << (t + 1 >> 1) * i;
  }
  printf("%d %d\n", k + !f[k], ans2);
  return 0;
}

```