#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 9;
bitset<N> bs;
int cnt[N], a[N], cst[N], arr[N], sz2 = 0, arr2[N], sz = 0;
bool vst[N];
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int n, k, i, j;
  cin >> n >> k;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) {
    if (!vst[i]) {
      vst[i]++;
      int len = 1, x = i;
      while (!vst[a[x]]) x = a[x], len++, vst[x]++;
      arr2[sz++] = len;
      cnt[len]++;
    }
  }
  for (i = 1; i <= n; i++) {
    if (cnt[i]) {
      int val = cnt[i];
      for (j = 0; (1 << j) <= val; val -= (1 << j), j++)
        arr[sz2++] = (1 << j) * i;
      if (val) arr[sz2++] = val * i;
    }
  }
  bs.set(0);
  for (i = 0; i < sz2; i++) bs |= bs << arr[i];
  cout << (bs.test(k) ? k : k + 1) << " ";
  memset(vst, 0, sizeof vst);
  int ans = 0, rmn = n;
  for (i = 0; i < sz; i++) {
    if (k * 2 >= arr2[i]) {
      k -= arr2[i] / 2;
      ans += arr2[i] - (arr2[i] & 1);
      arr2[i] &= 1;
    } else {
      ans += k * 2;
      k = 0;
    }
  }
  for (i = 0; i < sz && k; i++)
    if (arr2[i]) k--, ans++;
  cout << ans;
}

### 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 = 1e6 + 9;
bitset<N> bs;
int cnt[N], a[N], cst[N], arr[N], sz2 = 0, arr2[N], sz = 0;
bool vst[N];
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int n, k, i, j;
  cin >> n >> k;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) {
    if (!vst[i]) {
      vst[i]++;
      int len = 1, x = i;
      while (!vst[a[x]]) x = a[x], len++, vst[x]++;
      arr2[sz++] = len;
      cnt[len]++;
    }
  }
  for (i = 1; i <= n; i++) {
    if (cnt[i]) {
      int val = cnt[i];
      for (j = 0; (1 << j) <= val; val -= (1 << j), j++)
        arr[sz2++] = (1 << j) * i;
      if (val) arr[sz2++] = val * i;
    }
  }
  bs.set(0);
  for (i = 0; i < sz2; i++) bs |= bs << arr[i];
  cout << (bs.test(k) ? k : k + 1) << " ";
  memset(vst, 0, sizeof vst);
  int ans = 0, rmn = n;
  for (i = 0; i < sz; i++) {
    if (k * 2 >= arr2[i]) {
      k -= arr2[i] / 2;
      ans += arr2[i] - (arr2[i] & 1);
      arr2[i] &= 1;
    } else {
      ans += k * 2;
      k = 0;
    }
  }
  for (i = 0; i < sz && k; i++)
    if (arr2[i]) k--, ans++;
  cout << ans;
}

```