#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000005;
const int T = 100;
int N, K;
int p[MAXN];
bool bio[MAXN];
bitset<MAXN> dp;
int need[MAXN];
int cnt[T];
void load() {
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; i++) scanf("%d", p + i);
}
int dfs(int x) {
  if (bio[x]) return 0;
  bio[x] = true;
  return dfs(p[x]) + 1;
}
void solve() {
  dp.set(0);
  int sum = 0;
  for (int i = 1; i <= N; i++) {
    int tmp = dfs(i);
    if (tmp < T)
      cnt[tmp]++;
    else
      dp |= dp << tmp;
    sum += tmp / 2;
  }
  for (int i = 1; i < T; i++)
    for (int j = 0; j <= K; j++) {
      need[j] = MAXN;
      if (dp.test(j))
        need[j] = 0;
      else if (j >= i)
        need[j] = need[j - i] + 1;
      dp.set(j, need[j] <= cnt[i]);
    }
  printf("%d %d\n", K + !dp.test(K), min(N, K + min(K, sum)));
}
int main() {
  load();
  solve();
  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 int MAXN = 1000005;
const int T = 100;
int N, K;
int p[MAXN];
bool bio[MAXN];
bitset<MAXN> dp;
int need[MAXN];
int cnt[T];
void load() {
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; i++) scanf("%d", p + i);
}
int dfs(int x) {
  if (bio[x]) return 0;
  bio[x] = true;
  return dfs(p[x]) + 1;
}
void solve() {
  dp.set(0);
  int sum = 0;
  for (int i = 1; i <= N; i++) {
    int tmp = dfs(i);
    if (tmp < T)
      cnt[tmp]++;
    else
      dp |= dp << tmp;
    sum += tmp / 2;
  }
  for (int i = 1; i < T; i++)
    for (int j = 0; j <= K; j++) {
      need[j] = MAXN;
      if (dp.test(j))
        need[j] = 0;
      else if (j >= i)
        need[j] = need[j - i] + 1;
      dp.set(j, need[j] <= cnt[i]);
    }
  printf("%d %d\n", K + !dp.test(K), min(N, K + min(K, sum)));
}
int main() {
  load();
  solve();
  return 0;
}

```