#include <bits/stdc++.h>
using namespace std;
void read(int& x) { scanf("%d", &x); }
template <typename T, typename... Args>
void read(T& a, Args&... b) {
  read(a);
  read(b...);
}
const int max_n = int(1e6) + 10;
int n, k;
int nxt[max_n];
int vis[max_n];
int cnt[max_n];
bitset<max_n> dp;
int main() {
  read(n, k);
  for (int i = 1; i <= n; ++i) read(nxt[i]);
  vector<int> cs;
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) {
      int r = 1;
      vis[i] = 1;
      for (int j = nxt[i]; j != i; j = nxt[j]) {
        ++r;
        vis[j] = 1;
      }
      cs.push_back(r);
      ++cnt[r];
    }
  sort((cs).begin(), (cs).end());
  dp[0] = 1;
  for (int i = 1; i <= n; ++i) {
    if (!cnt[i]) continue;
    for (int j = 0; j <= 20; ++j) {
      if ((1 << j) <= cnt[i]) dp |= (dp << (i * (1 << j))), cnt[i] -= (1 << j);
    }
    if (cnt[i]) dp |= (dp << (i * cnt[i]));
  }
  printf("%d ", k + 1 - dp[k]);
  int mc = 0;
  for (int x : cs) mc += x / 2;
  if (k <= mc) {
    printf("%d\n", k * 2);
  } else {
    printf("%d\n", min(n, k + mc));
  }
  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;
void read(int& x) { scanf("%d", &x); }
template <typename T, typename... Args>
void read(T& a, Args&... b) {
  read(a);
  read(b...);
}
const int max_n = int(1e6) + 10;
int n, k;
int nxt[max_n];
int vis[max_n];
int cnt[max_n];
bitset<max_n> dp;
int main() {
  read(n, k);
  for (int i = 1; i <= n; ++i) read(nxt[i]);
  vector<int> cs;
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) {
      int r = 1;
      vis[i] = 1;
      for (int j = nxt[i]; j != i; j = nxt[j]) {
        ++r;
        vis[j] = 1;
      }
      cs.push_back(r);
      ++cnt[r];
    }
  sort((cs).begin(), (cs).end());
  dp[0] = 1;
  for (int i = 1; i <= n; ++i) {
    if (!cnt[i]) continue;
    for (int j = 0; j <= 20; ++j) {
      if ((1 << j) <= cnt[i]) dp |= (dp << (i * (1 << j))), cnt[i] -= (1 << j);
    }
    if (cnt[i]) dp |= (dp << (i * cnt[i]));
  }
  printf("%d ", k + 1 - dp[k]);
  int mc = 0;
  for (int x : cs) mc += x / 2;
  if (k <= mc) {
    printf("%d\n", k * 2);
  } else {
    printf("%d\n", min(n, k + mc));
  }
  return 0;
}

```