#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000001;
int n, k, x, cnt[maxn];
bool vis[maxn];
vector<int> a;
bitset<maxn> dp;
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n && scanf("%d", &x); i++) a.push_back(x - 1);
  int tot = 0;
  for (int i = 0; i < n; i++) {
    if (vis[i]) continue;
    int len = 0, cur = i;
    while (vis[cur] == 0) {
      len++;
      vis[cur] = 1;
      cur = a[cur];
    }
    tot += len / 2;
    cnt[len]++;
  }
  dp.set(0);
  for (int i = 1; i <= n; i++) {
    if (cnt[i] == 0) continue;
    for (int j = 0; (1 << j) <= cnt[i]; j++) {
      dp |= (dp << (i << j));
      cnt[i] -= (1 << j);
    }
    dp |= (dp << (i * cnt[i]));
  }
  printf("%d ", k + !dp[k]);
  if (k <= tot)
    printf("%d\n", 2 * k);
  else
    printf("%d\n", min(n, tot + k));
  return 0;
}

### Prompt

Your task is to create 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;
const int maxn = 1000001;
int n, k, x, cnt[maxn];
bool vis[maxn];
vector<int> a;
bitset<maxn> dp;
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n && scanf("%d", &x); i++) a.push_back(x - 1);
  int tot = 0;
  for (int i = 0; i < n; i++) {
    if (vis[i]) continue;
    int len = 0, cur = i;
    while (vis[cur] == 0) {
      len++;
      vis[cur] = 1;
      cur = a[cur];
    }
    tot += len / 2;
    cnt[len]++;
  }
  dp.set(0);
  for (int i = 1; i <= n; i++) {
    if (cnt[i] == 0) continue;
    for (int j = 0; (1 << j) <= cnt[i]; j++) {
      dp |= (dp << (i << j));
      cnt[i] -= (1 << j);
    }
    dp |= (dp << (i * cnt[i]));
  }
  printf("%d ", k + !dp[k]);
  if (k <= tot)
    printf("%d\n", 2 * k);
  else
    printf("%d\n", min(n, tot + k));
  return 0;
}

```