#include <bits/stdc++.h>
using namespace std;
int p[1100000];
bool d[1100000];
int kubelki[1100000];
bitset<1100000> dp;
int last[1100000];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &p[i]);
    --p[i];
  }
  vector<int> V;
  for (int i = 0; i < n; ++i) {
    if (d[i] == false) {
      int s = p[i];
      int ile = 1;
      while (s != i) {
        d[s] = true;
        ++ile;
        s = p[s];
      }
      ++kubelki[ile];
    }
  }
  dp[0] = true;
  for (int i = k; i > 100; --i)
    while (kubelki[i]) {
      dp |= (dp << i);
      --kubelki[i];
    }
  for (int i = min(100, k); i > 0 && dp[k] == false; --i)
    if (kubelki[i]) {
      for (int j = 0; j < i; ++j) last[j] = 1000000000;
      last[0] = 1;
      int rest = -1;
      for (int j = i; j <= k; ++j) {
        ++rest;
        if (rest == i) rest = 0;
        if (dp[j])
          last[rest] = 0;
        else if (last[rest] <= kubelki[i])
          dp[j] = true;
        ++last[rest];
      }
    }
  int ilenp = 0;
  for (int i = 1; i <= n; i += 2) ilenp += kubelki[i];
  int ilepodwojnych = (n - ilenp) / 2;
  int resmax = 0;
  if (k < ilepodwojnych)
    resmax = 2 * k;
  else
    resmax = min(n, k + ilepodwojnych);
  int resmin = (dp[k] ? k : k + 1);
  printf("%d %d\n", resmin, resmax);
}

### 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;
int p[1100000];
bool d[1100000];
int kubelki[1100000];
bitset<1100000> dp;
int last[1100000];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &p[i]);
    --p[i];
  }
  vector<int> V;
  for (int i = 0; i < n; ++i) {
    if (d[i] == false) {
      int s = p[i];
      int ile = 1;
      while (s != i) {
        d[s] = true;
        ++ile;
        s = p[s];
      }
      ++kubelki[ile];
    }
  }
  dp[0] = true;
  for (int i = k; i > 100; --i)
    while (kubelki[i]) {
      dp |= (dp << i);
      --kubelki[i];
    }
  for (int i = min(100, k); i > 0 && dp[k] == false; --i)
    if (kubelki[i]) {
      for (int j = 0; j < i; ++j) last[j] = 1000000000;
      last[0] = 1;
      int rest = -1;
      for (int j = i; j <= k; ++j) {
        ++rest;
        if (rest == i) rest = 0;
        if (dp[j])
          last[rest] = 0;
        else if (last[rest] <= kubelki[i])
          dp[j] = true;
        ++last[rest];
      }
    }
  int ilenp = 0;
  for (int i = 1; i <= n; i += 2) ilenp += kubelki[i];
  int ilepodwojnych = (n - ilenp) / 2;
  int resmax = 0;
  if (k < ilepodwojnych)
    resmax = 2 * k;
  else
    resmax = min(n, k + ilepodwojnych);
  int resmin = (dp[k] ? k : k + 1);
  printf("%d %d\n", resmin, resmax);
}

```