#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 6;
int mn;
int mx;
int n, k;
int nxt[N];
bool visited[N];
bitset<N> dp;
int cnt[N];
int arr[N];
int sz;
int arr2[N];
int sz2;
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", nxt + i);
  }
  sz = 0;
  mn = 0;
  mx = 0;
  for (int i = 1; i <= n; ++i) {
    visited[i] = 0;
    cnt[i] = 0;
  }
  for (int i = 1; i <= n; ++i) {
    if (!visited[i]) {
      int node = i;
      int cnt = 0;
      do {
        visited[node] = 1;
        node = nxt[node];
        ++cnt;
      } while (node != i);
      ++::cnt[cnt];
      arr[++sz] = cnt;
    }
  }
  sz2 = 0;
  for (int i = 1; i <= n; ++i) {
    if (cnt[i]) {
      int val = cnt[i];
      for (int j = 0; (1 << j) <= val; ++j) {
        arr2[++sz2] = (1 << j) * i;
        val -= 1 << j;
      }
      if (val) {
        arr2[++sz2] = val * i;
      }
    }
  }
  dp.set(0);
  for (int i = 1; i <= sz2; ++i) {
    dp |= dp << arr2[i];
  }
  if (dp.test(k)) {
    mn = k;
  } else {
    mn = k + 1;
  }
  for (int i = 1; i <= sz && k; ++i) {
    if (k * 2 >= arr[i]) {
      k -= arr[i] >> 1;
      mx += arr[i] - (arr[i] & 1);
      arr[i] &= 1;
    } else {
      mx += k * 2;
      arr[i] -= k * 2;
      k = 0;
    }
  }
  for (int i = 1; i <= sz && k; ++i) {
    if (arr[i]) {
      --k;
      --arr[i];
      ++mx;
    }
  }
  printf("%d %d\n", mn, mx);
}

### 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 + 6;
int mn;
int mx;
int n, k;
int nxt[N];
bool visited[N];
bitset<N> dp;
int cnt[N];
int arr[N];
int sz;
int arr2[N];
int sz2;
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", nxt + i);
  }
  sz = 0;
  mn = 0;
  mx = 0;
  for (int i = 1; i <= n; ++i) {
    visited[i] = 0;
    cnt[i] = 0;
  }
  for (int i = 1; i <= n; ++i) {
    if (!visited[i]) {
      int node = i;
      int cnt = 0;
      do {
        visited[node] = 1;
        node = nxt[node];
        ++cnt;
      } while (node != i);
      ++::cnt[cnt];
      arr[++sz] = cnt;
    }
  }
  sz2 = 0;
  for (int i = 1; i <= n; ++i) {
    if (cnt[i]) {
      int val = cnt[i];
      for (int j = 0; (1 << j) <= val; ++j) {
        arr2[++sz2] = (1 << j) * i;
        val -= 1 << j;
      }
      if (val) {
        arr2[++sz2] = val * i;
      }
    }
  }
  dp.set(0);
  for (int i = 1; i <= sz2; ++i) {
    dp |= dp << arr2[i];
  }
  if (dp.test(k)) {
    mn = k;
  } else {
    mn = k + 1;
  }
  for (int i = 1; i <= sz && k; ++i) {
    if (k * 2 >= arr[i]) {
      k -= arr[i] >> 1;
      mx += arr[i] - (arr[i] & 1);
      arr[i] &= 1;
    } else {
      mx += k * 2;
      arr[i] -= k * 2;
      k = 0;
    }
  }
  for (int i = 1; i <= sz && k; ++i) {
    if (arr[i]) {
      --k;
      --arr[i];
      ++mx;
    }
  }
  printf("%d %d\n", mn, mx);
}

```