#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 5, T = 100;
int n, a[MAXN], k, cnt = 0, t[MAXN];
bool visit[MAXN];
bitset<MAXN> btset;
void dfs(int v) {
  visit[v] = true;
  cnt++;
  if (!visit[a[v]]) dfs(a[v]);
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  int kk = 0, o = 0;
  for (int i = 0; i < n; i++)
    if (!visit[i]) {
      cnt = 0;
      dfs(i);
      kk += (cnt / 2);
      o += (cnt % 2);
      t[cnt]++;
    }
  btset[0] = 1;
  for (int i = T; i <= n; i++)
    for (int j = 0; j < t[i]; j++) btset |= (btset << i);
  for (int i = 1; i < T; i++) {
    if (t[i] == 0) continue;
    for (int j = n; j >= 0; j--) {
      if (btset[j] == 0) continue;
      int ii = j, c = t[i];
      while (ii + i <= n && c > 0 && !btset[ii + i]) {
        ii += i;
        btset[ii] = 1;
        c--;
      }
    }
  }
  int mn = ((btset[k]) ? k : k + 1),
      mx = ((kk >= k) ? 2 * k : (2 * kk) + min(k - kk, o));
  cout << mn << " " << mx << endl;
  return 0;
}

### Prompt

Please formulate 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 = 1e6 + 5, T = 100;
int n, a[MAXN], k, cnt = 0, t[MAXN];
bool visit[MAXN];
bitset<MAXN> btset;
void dfs(int v) {
  visit[v] = true;
  cnt++;
  if (!visit[a[v]]) dfs(a[v]);
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  int kk = 0, o = 0;
  for (int i = 0; i < n; i++)
    if (!visit[i]) {
      cnt = 0;
      dfs(i);
      kk += (cnt / 2);
      o += (cnt % 2);
      t[cnt]++;
    }
  btset[0] = 1;
  for (int i = T; i <= n; i++)
    for (int j = 0; j < t[i]; j++) btset |= (btset << i);
  for (int i = 1; i < T; i++) {
    if (t[i] == 0) continue;
    for (int j = n; j >= 0; j--) {
      if (btset[j] == 0) continue;
      int ii = j, c = t[i];
      while (ii + i <= n && c > 0 && !btset[ii + i]) {
        ii += i;
        btset[ii] = 1;
        c--;
      }
    }
  }
  int mn = ((btset[k]) ? k : k + 1),
      mx = ((kk >= k) ? 2 * k : (2 * kk) + min(k - kk, o));
  cout << mn << " " << mx << endl;
  return 0;
}

```