#include <bits/stdc++.h>
using namespace std;
int n, k, p[1000003], cntt[74];
bool vis[1000003];
vector<int> v;
int dfs(int x) {
  vis[x] = 1;
  if (vis[p[x]]) return 1;
  return dfs(p[x]) + 1;
}
int read() {
  int ret = 0;
  char c;
  while (c > '9' || c < '0') c = getchar();
  while ('0' <= c && c <= '9') ret = ret * 10 + c - '0', c = getchar();
  return ret;
}
bitset<1000003> x;
int main() {
  n = read();
  k = read();
  for (int i = 1; i <= n; i++) p[i] = read();
  x[0] = 1;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) {
      int tmp = dfs(i);
      v.push_back(tmp);
      if (tmp >= 73)
        x |= (x << tmp);
      else
        cntt[tmp]++;
    }
  for (int i = 1; i <= 73; i++)
    if (cntt[i]) {
      int j = 1;
      for (; j * 2 - 1 <= cntt[i]; j *= 2) x |= (x << j * i);
      j /= 2;
      x |= (x << (cntt[i] - j * 2 + 1) * i);
    }
  if (x[k])
    cout << k << ' ';
  else
    cout << k + 1 << ' ';
  int cnt = 0;
  for (int i = 0; i < v.size(); i++) cnt += v[i] / 2;
  if (k < cnt) {
    cout << k * 2;
    return 0;
  }
  cout << min(cnt + k, n);
}

### Prompt

Your challenge is to write a CPP solution 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 n, k, p[1000003], cntt[74];
bool vis[1000003];
vector<int> v;
int dfs(int x) {
  vis[x] = 1;
  if (vis[p[x]]) return 1;
  return dfs(p[x]) + 1;
}
int read() {
  int ret = 0;
  char c;
  while (c > '9' || c < '0') c = getchar();
  while ('0' <= c && c <= '9') ret = ret * 10 + c - '0', c = getchar();
  return ret;
}
bitset<1000003> x;
int main() {
  n = read();
  k = read();
  for (int i = 1; i <= n; i++) p[i] = read();
  x[0] = 1;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) {
      int tmp = dfs(i);
      v.push_back(tmp);
      if (tmp >= 73)
        x |= (x << tmp);
      else
        cntt[tmp]++;
    }
  for (int i = 1; i <= 73; i++)
    if (cntt[i]) {
      int j = 1;
      for (; j * 2 - 1 <= cntt[i]; j *= 2) x |= (x << j * i);
      j /= 2;
      x |= (x << (cntt[i] - j * 2 + 1) * i);
    }
  if (x[k])
    cout << k << ' ';
  else
    cout << k + 1 << ' ';
  int cnt = 0;
  for (int i = 0; i < v.size(); i++) cnt += v[i] / 2;
  if (k < cnt) {
    cout << k * 2;
    return 0;
  }
  cout << min(cnt + k, n);
}

```