#include <bits/stdc++.h>
using namespace std;
const unsigned int max_n = 1000005;
int FUCK_STD[max_n];
bool vis[max_n];
bitset<max_n> b1, b2;
vector<int> sz;
vector<int> ruc;
void add(int a, int num) {
  int curs = 1;
  while (curs <= num) {
    ruc.push_back(curs * a);
    num -= curs;
    curs *= 2;
  }
  if (num) {
    ruc.push_back(a * num);
  }
}
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    scanf("%d", FUCK_STD + i);
    FUCK_STD[i]--;
  }
  int len, cur;
  for (int i = 0; i < n; ++i) {
    if (!vis[i]) {
      vis[i] = true;
      cur = FUCK_STD[i];
      len = 1;
      while (!vis[cur]) {
        len++;
        vis[cur] = true;
        cur = FUCK_STD[cur];
      }
      sz.push_back(len);
    }
  }
  sort(sz.begin(), sz.end());
  int num = 1;
  for (int i = 1; i < sz.size(); ++i) {
    if (sz[i] == sz[i - 1]) {
      num++;
    } else {
      add(sz[i - 1], num);
      num = 1;
    }
  }
  add(sz[sz.size() - 1], num);
  b1[0] = true;
  for (int i = 0; i < ruc.size(); ++i) {
    b2 = (b1 << ruc[i]);
    b1 |= b2;
  }
  if (b1[k]) {
    cout << k << ' ';
  } else {
    cout << k + 1 << ' ';
  }
  int mans = 0;
  for (int i = 0; i < sz.size(); ++i) {
    if (k == 0) {
      break;
    }
    if (sz[i] / 2 < k) {
      k -= sz[i] / 2;
      mans += (sz[i] / 2) * 2;
    } else {
      mans += k * 2;
      k = 0;
    }
  }
  for (int i = 0; i < sz.size(); ++i) {
    if (k == 0) {
      break;
    }
    if (sz[i] % 2 == 1) {
      k--;
      mans++;
    }
  }
  cout << mans;
  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 unsigned int max_n = 1000005;
int FUCK_STD[max_n];
bool vis[max_n];
bitset<max_n> b1, b2;
vector<int> sz;
vector<int> ruc;
void add(int a, int num) {
  int curs = 1;
  while (curs <= num) {
    ruc.push_back(curs * a);
    num -= curs;
    curs *= 2;
  }
  if (num) {
    ruc.push_back(a * num);
  }
}
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    scanf("%d", FUCK_STD + i);
    FUCK_STD[i]--;
  }
  int len, cur;
  for (int i = 0; i < n; ++i) {
    if (!vis[i]) {
      vis[i] = true;
      cur = FUCK_STD[i];
      len = 1;
      while (!vis[cur]) {
        len++;
        vis[cur] = true;
        cur = FUCK_STD[cur];
      }
      sz.push_back(len);
    }
  }
  sort(sz.begin(), sz.end());
  int num = 1;
  for (int i = 1; i < sz.size(); ++i) {
    if (sz[i] == sz[i - 1]) {
      num++;
    } else {
      add(sz[i - 1], num);
      num = 1;
    }
  }
  add(sz[sz.size() - 1], num);
  b1[0] = true;
  for (int i = 0; i < ruc.size(); ++i) {
    b2 = (b1 << ruc[i]);
    b1 |= b2;
  }
  if (b1[k]) {
    cout << k << ' ';
  } else {
    cout << k + 1 << ' ';
  }
  int mans = 0;
  for (int i = 0; i < sz.size(); ++i) {
    if (k == 0) {
      break;
    }
    if (sz[i] / 2 < k) {
      k -= sz[i] / 2;
      mans += (sz[i] / 2) * 2;
    } else {
      mans += k * 2;
      k = 0;
    }
  }
  for (int i = 0; i < sz.size(); ++i) {
    if (k == 0) {
      break;
    }
    if (sz[i] % 2 == 1) {
      k--;
      mans++;
    }
  }
  cout << mans;
  return 0;
}

```