#include <bits/stdc++.h>
using namespace std;
const int N = 1000010;
bitset<N> dp;
int p[N], A[N], vis[N];
long long buc[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1, n, i, j, k;
  cin >> n >> k;
  for (i = 1; i <= n; i++) cin >> p[i];
  int len = 0;
  for (i = 1; i <= n; i++) {
    if (!vis[i]) {
      int tp = i;
      int c = 0;
      do {
        vis[tp] = 1;
        tp = p[tp];
        c++;
      } while (tp != i);
      buc[c]++;
    }
  }
  for (i = 1; i <= n; i++) {
    if (buc[i]) {
      int val = buc[i];
      int j = 0;
      long long tot = 0;
      while ((1 << j) <= val) {
        A[++len] = i * (1 << j);
        val -= (1 << j);
        j++;
        tot += A[len];
      }
      if (tot != 1ll * buc[i] * i) A[++len] = (1ll * i * buc[i] - tot);
    }
  }
  int mx;
  dp.set(0);
  for (i = 1; i <= len; i++) dp |= (dp << A[i]);
  if (dp.test(k))
    mx = k;
  else
    mx = k + 1;
  int temp = k;
  for (i = 1; i <= n; i++) temp -= (buc[i] * (i / 2));
  temp = max(temp, 0);
  int ans = 2 * (k - temp);
  if (temp) {
    ans += temp;
  }
  cout << min(mx, n) << " " << min(n, ans) << "\n";
  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;
const int N = 1000010;
bitset<N> dp;
int p[N], A[N], vis[N];
long long buc[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1, n, i, j, k;
  cin >> n >> k;
  for (i = 1; i <= n; i++) cin >> p[i];
  int len = 0;
  for (i = 1; i <= n; i++) {
    if (!vis[i]) {
      int tp = i;
      int c = 0;
      do {
        vis[tp] = 1;
        tp = p[tp];
        c++;
      } while (tp != i);
      buc[c]++;
    }
  }
  for (i = 1; i <= n; i++) {
    if (buc[i]) {
      int val = buc[i];
      int j = 0;
      long long tot = 0;
      while ((1 << j) <= val) {
        A[++len] = i * (1 << j);
        val -= (1 << j);
        j++;
        tot += A[len];
      }
      if (tot != 1ll * buc[i] * i) A[++len] = (1ll * i * buc[i] - tot);
    }
  }
  int mx;
  dp.set(0);
  for (i = 1; i <= len; i++) dp |= (dp << A[i]);
  if (dp.test(k))
    mx = k;
  else
    mx = k + 1;
  int temp = k;
  for (i = 1; i <= n; i++) temp -= (buc[i] * (i / 2));
  temp = max(temp, 0);
  int ans = 2 * (k - temp);
  if (temp) {
    ans += temp;
  }
  cout << min(mx, n) << " " << min(n, ans) << "\n";
  return 0;
}

```