#include <bits/stdc++.h>
using namespace std;
template <class T>
inline int size(const T& t) {
  return t.size();
}
int N, K;
int perm[1000000];
vector<int> cykle;
bool tab[1000001];
int czy() {
  if (K + K > N) K = N - K;
  if (!K) return 1;
  int sum = 0;
  tab[0] = 1;
  sort(cykle.begin(), cykle.end());
  for (int a = 0; a < (size(cykle)); ++a) {
    int len = cykle[a];
    bool powt = a && cykle[a] == cykle[a - 1];
    for (int b = (min(sum, K - len)); b >= (0); --b) {
      if (tab[b]) tab[b + len] = 1;
      if (powt && b <= sum - len) break;
    }
    if (tab[K]) return 1;
    sum += len;
  }
  return 0;
}
int main() {
  scanf("%d%d", &N, &K);
  for (int a = 0; a < (N); ++a) {
    scanf("%d", &perm[a]);
    --perm[a];
  }
  int ss = 0;
  for (int a = 0; a < (N); ++a)
    if (perm[a] >= 0) {
      int x = a;
      int len = 0;
      while (perm[x] >= 0) {
        int n = perm[x];
        perm[x] = -1;
        x = n;
        ++len;
      }
      cykle.push_back(len);
      ss += len / 2;
    }
  int ile2 = min(K, ss);
  int w2 = min(ile2 * 2 + K - ile2, N);
  int w1 = K;
  printf("%d %d\n", w1 + 1 - czy(), w2);
}

### 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;
template <class T>
inline int size(const T& t) {
  return t.size();
}
int N, K;
int perm[1000000];
vector<int> cykle;
bool tab[1000001];
int czy() {
  if (K + K > N) K = N - K;
  if (!K) return 1;
  int sum = 0;
  tab[0] = 1;
  sort(cykle.begin(), cykle.end());
  for (int a = 0; a < (size(cykle)); ++a) {
    int len = cykle[a];
    bool powt = a && cykle[a] == cykle[a - 1];
    for (int b = (min(sum, K - len)); b >= (0); --b) {
      if (tab[b]) tab[b + len] = 1;
      if (powt && b <= sum - len) break;
    }
    if (tab[K]) return 1;
    sum += len;
  }
  return 0;
}
int main() {
  scanf("%d%d", &N, &K);
  for (int a = 0; a < (N); ++a) {
    scanf("%d", &perm[a]);
    --perm[a];
  }
  int ss = 0;
  for (int a = 0; a < (N); ++a)
    if (perm[a] >= 0) {
      int x = a;
      int len = 0;
      while (perm[x] >= 0) {
        int n = perm[x];
        perm[x] = -1;
        x = n;
        ++len;
      }
      cykle.push_back(len);
      ss += len / 2;
    }
  int ile2 = min(K, ss);
  int w2 = min(ile2 * 2 + K - ile2, N);
  int w1 = K;
  printf("%d %d\n", w1 + 1 - czy(), w2);
}

```