#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
map<int, int> mp;
int p[maxn];
bool vis[maxn];
bitset<maxn> f;
int main() {
  int n, k, i, j, l, s1, s2, a, b;
  s1 = s2 = 0;
  scanf("%d%d", &n, &k);
  for (i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (i = 1; i <= n; i++) {
    if (!vis[i]) {
      j = p[i];
      l = 1;
      while (j != i) {
        vis[j] = 1;
        j = p[j];
        l++;
      }
      mp[l]++;
      s2 += l / 2;
    }
  }
  if (s2 >= k)
    s2 = k * 2;
  else
    s2 += k;
  s2 = min(n, s2);
  f[0] = 1;
  for (auto u : mp) {
    a = u.first;
    b = u.second;
    j = 0;
    for (i = 0; j + (1 << i) < b; i++) {
      f |= f << (a * (1 << i));
      j |= 1 << i;
    }
    b -= j;
    f |= f << (a * b);
    if (f[k]) break;
  }
  s1 = k + (f[k] != 1);
  printf("%d %d\n", s1, s2);
  return 0;
}

### Prompt

Please create a solution in Cpp 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 maxn = 1e6 + 5;
map<int, int> mp;
int p[maxn];
bool vis[maxn];
bitset<maxn> f;
int main() {
  int n, k, i, j, l, s1, s2, a, b;
  s1 = s2 = 0;
  scanf("%d%d", &n, &k);
  for (i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (i = 1; i <= n; i++) {
    if (!vis[i]) {
      j = p[i];
      l = 1;
      while (j != i) {
        vis[j] = 1;
        j = p[j];
        l++;
      }
      mp[l]++;
      s2 += l / 2;
    }
  }
  if (s2 >= k)
    s2 = k * 2;
  else
    s2 += k;
  s2 = min(n, s2);
  f[0] = 1;
  for (auto u : mp) {
    a = u.first;
    b = u.second;
    j = 0;
    for (i = 0; j + (1 << i) < b; i++) {
      f |= f << (a * (1 << i));
      j |= 1 << i;
    }
    b -= j;
    f |= f << (a * b);
    if (f[k]) break;
  }
  s1 = k + (f[k] != 1);
  printf("%d %d\n", s1, s2);
  return 0;
}

```