#include <bits/stdc++.h>
using namespace std;
bool used[1000010];
int n, k;
int nums[1000010];
int size = 0;
int datas[1000010];
bool pack[1000010];
int main() {
  scanf("%d %d", &n, &k);
  int i;
  for (i = 1; i <= n; i++) scanf("%d", &datas[i]);
  for (i = 1; i <= n; i++) {
    if (used[i]) continue;
    used[i] = true;
    nums[size] = 1;
    int k = datas[i];
    while (!used[k]) {
      used[k] = true;
      nums[size]++;
      k = datas[k];
    }
    size++;
  }
  map<int, int> count;
  for (i = 0; i < size; i++) {
    count[nums[i]] += 1;
  }
  bitset<1000010> b;
  b[0] = 1;
  map<int, int>::iterator it = count.begin();
  for (; it != count.end(); it++) {
    int a = 1;
    int amount = it->second;
    while (a < amount) {
      amount -= a;
      b |= b << a * it->first;
      a *= 2;
    }
    b |= b << amount * it->first;
  }
  if (b[k])
    printf("%d ", k);
  else
    printf("%d ", k + 1);
  int oc = 0;
  int k_ = k;
  for (i = 0; i < size; i++) {
    if (nums[i] % 2 == 0 && k_ >= nums[i] / 2)
      k_ -= nums[i] / 2;
    else if (k_ >= nums[i] / 2) {
      {
        k_ -= nums[i] / 2;
        oc++;
      }
    } else {
      k_ = 0;
      break;
    }
  }
  if (i == size && k_ >= 0) {
    if (k_ >= oc)
      printf("%d\n", n);
    else
      printf("%d\n", 2 * (k - k_) + k_);
  } else
    printf("%d\n", 2 * k);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
bool used[1000010];
int n, k;
int nums[1000010];
int size = 0;
int datas[1000010];
bool pack[1000010];
int main() {
  scanf("%d %d", &n, &k);
  int i;
  for (i = 1; i <= n; i++) scanf("%d", &datas[i]);
  for (i = 1; i <= n; i++) {
    if (used[i]) continue;
    used[i] = true;
    nums[size] = 1;
    int k = datas[i];
    while (!used[k]) {
      used[k] = true;
      nums[size]++;
      k = datas[k];
    }
    size++;
  }
  map<int, int> count;
  for (i = 0; i < size; i++) {
    count[nums[i]] += 1;
  }
  bitset<1000010> b;
  b[0] = 1;
  map<int, int>::iterator it = count.begin();
  for (; it != count.end(); it++) {
    int a = 1;
    int amount = it->second;
    while (a < amount) {
      amount -= a;
      b |= b << a * it->first;
      a *= 2;
    }
    b |= b << amount * it->first;
  }
  if (b[k])
    printf("%d ", k);
  else
    printf("%d ", k + 1);
  int oc = 0;
  int k_ = k;
  for (i = 0; i < size; i++) {
    if (nums[i] % 2 == 0 && k_ >= nums[i] / 2)
      k_ -= nums[i] / 2;
    else if (k_ >= nums[i] / 2) {
      {
        k_ -= nums[i] / 2;
        oc++;
      }
    } else {
      k_ = 0;
      break;
    }
  }
  if (i == size && k_ >= 0) {
    if (k_ >= oc)
      printf("%d\n", n);
    else
      printf("%d\n", 2 * (k - k_) + k_);
  } else
    printf("%d\n", 2 * k);
  return 0;
}

```