#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 5;
int n, p[MAXN], k;
vector<int> l;
bool vis[MAXN];
int even = 0, odd = 0;
int ct1[MAXN];
bitset<MAXN> b;
int num[MAXN];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", p + i);
    p[i]--;
  }
  for (int i = 0; i < n; ++i) {
    if (!vis[i]) {
      int ct = 0;
      int j = i;
      while (!vis[j]) {
        vis[j] = 1;
        ct++;
        j = p[j];
      }
      ct1[ct]++;
      if (ct % 2 == 0)
        even++;
      else
        odd++;
    }
  }
  b[0] = 1;
  for (int i = 1; i <= n; ++i) {
    if (ct1[i] == 0) continue;
    if (i <= 197) {
      memset(num, 0, sizeof(num));
      for (int j = i; j <= n; ++j) {
        if (b[j] || (num[j - i] == ct1[i]) || (!b[j - i])) continue;
        b[j] = true;
        num[j] = num[j - i] + 1;
      }
    } else {
      for (int j = 0; j < ct1[i]; ++j) b |= (b << i);
    }
  }
  int mini = 0;
  if (b[k])
    mini = k;
  else
    mini = k + 1;
  int maxi = 0;
  if (k <= (n - odd) / 2)
    maxi = 2 * k;
  else
    maxi = min(n, (n - odd) + (k - (n - odd) / 2));
  cout << mini << ' ' << maxi << endl;
}

### 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 + 5;
int n, p[MAXN], k;
vector<int> l;
bool vis[MAXN];
int even = 0, odd = 0;
int ct1[MAXN];
bitset<MAXN> b;
int num[MAXN];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", p + i);
    p[i]--;
  }
  for (int i = 0; i < n; ++i) {
    if (!vis[i]) {
      int ct = 0;
      int j = i;
      while (!vis[j]) {
        vis[j] = 1;
        ct++;
        j = p[j];
      }
      ct1[ct]++;
      if (ct % 2 == 0)
        even++;
      else
        odd++;
    }
  }
  b[0] = 1;
  for (int i = 1; i <= n; ++i) {
    if (ct1[i] == 0) continue;
    if (i <= 197) {
      memset(num, 0, sizeof(num));
      for (int j = i; j <= n; ++j) {
        if (b[j] || (num[j - i] == ct1[i]) || (!b[j - i])) continue;
        b[j] = true;
        num[j] = num[j - i] + 1;
      }
    } else {
      for (int j = 0; j < ct1[i]; ++j) b |= (b << i);
    }
  }
  int mini = 0;
  if (b[k])
    mini = k;
  else
    mini = k + 1;
  int maxi = 0;
  if (k <= (n - odd) / 2)
    maxi = 2 * k;
  else
    maxi = min(n, (n - odd) + (k - (n - odd) / 2));
  cout << mini << ' ' << maxi << endl;
}

```