#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const int nax = 1e6 + 5;
int cnt[nax], p[nax];
bool vis[nax];
bitset<nax> dp;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int seco = 0;
  for (int i = 1; i <= n; ++i) scanf("%d", &p[i]);
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) {
      int size = 0;
      do {
        vis[i] = true;
        ++size;
        i = p[i];
      } while (!vis[i]);
      seco += size / 2;
      ++cnt[size];
    }
  dp[0] = true;
  for (int size = 1; size <= n; ++size)
    if (cnt[size]) {
      int x = cnt[size];
      for (int i = 0; x; ++i) {
        dp |= dp << (size << i);
        x -= 1 << i;
        if (x & (1 << i)) {
          dp |= dp << (size << i);
          x -= 1 << i;
        }
      }
    }
  int strong = min(seco, k);
  printf("%d %d\n", k + !dp[k], min(n, strong * 2 + (k - strong)));
}

### Prompt

In CPP, your task is to solve 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 c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const int nax = 1e6 + 5;
int cnt[nax], p[nax];
bool vis[nax];
bitset<nax> dp;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int seco = 0;
  for (int i = 1; i <= n; ++i) scanf("%d", &p[i]);
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) {
      int size = 0;
      do {
        vis[i] = true;
        ++size;
        i = p[i];
      } while (!vis[i]);
      seco += size / 2;
      ++cnt[size];
    }
  dp[0] = true;
  for (int size = 1; size <= n; ++size)
    if (cnt[size]) {
      int x = cnt[size];
      for (int i = 0; x; ++i) {
        dp |= dp << (size << i);
        x -= 1 << i;
        if (x & (1 << i)) {
          dp |= dp << (size << i);
          x -= 1 << i;
        }
      }
    }
  int strong = min(seco, k);
  printf("%d %d\n", k + !dp[k], min(n, strong * 2 + (k - strong)));
}

```