#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000010;
int p[MAXN], cnt[MAXN];
bitset<MAXN> b;
bool past[MAXN];
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> p[i];
  }
  b.set(n);
  int min_ans = 0, max_ans = 0, nOdd = 0, Kneeded = 0;
  for (int i = 1; i <= n; ++i) {
    if (past[i]) continue;
    int x = i, len = 0;
    while (!past[x]) {
      len++;
      past[x] = true;
      x = p[x];
    }
    cnt[len]++;
    Kneeded += len / 2;
    nOdd += len % 2;
  }
  for (int i = 1; i <= n; ++i) {
    if (cnt[i] == 0) continue;
    int j = 1;
    while (cnt[i] > 0) {
      j = min(j, cnt[i]);
      cnt[i] -= j;
      int u = i * j;
      b |= b >> u;
      j *= 2;
    }
  }
  min_ans = k;
  if (!b.test(n - k)) min_ans++;
  max_ans += 2 * min(k, Kneeded);
  int tempk = k - min(k, Kneeded);
  max_ans += min(tempk, nOdd);
  cout << min_ans << " " << max_ans;
  return 0;
}

### Prompt

Your challenge is to write a cpp solution 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 = 1000010;
int p[MAXN], cnt[MAXN];
bitset<MAXN> b;
bool past[MAXN];
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> p[i];
  }
  b.set(n);
  int min_ans = 0, max_ans = 0, nOdd = 0, Kneeded = 0;
  for (int i = 1; i <= n; ++i) {
    if (past[i]) continue;
    int x = i, len = 0;
    while (!past[x]) {
      len++;
      past[x] = true;
      x = p[x];
    }
    cnt[len]++;
    Kneeded += len / 2;
    nOdd += len % 2;
  }
  for (int i = 1; i <= n; ++i) {
    if (cnt[i] == 0) continue;
    int j = 1;
    while (cnt[i] > 0) {
      j = min(j, cnt[i]);
      cnt[i] -= j;
      int u = i * j;
      b |= b >> u;
      j *= 2;
    }
  }
  min_ans = k;
  if (!b.test(n - k)) min_ans++;
  max_ans += 2 * min(k, Kneeded);
  int tempk = k - min(k, Kneeded);
  max_ans += min(tempk, nOdd);
  cout << min_ans << " " << max_ans;
  return 0;
}

```