#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10, mod = 1e9 + 7;
const long long inf = 1e18;
int f[maxn], cnt[maxn], cn1, sm, ans1, ans2;
bool mark[maxn];
bitset<maxn> ans;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &f[i]);
  for (int i = 1; i <= n; i++) {
    if (!mark[i]) {
      int tmp = i, ln = 0;
      while (!mark[tmp]) {
        mark[tmp] = 1;
        ln++;
        tmp = f[tmp];
      }
      sm += ln / 2;
      cn1 += ln & 1;
      cnt[ln]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    while (cnt[i] > 2) cnt[i] -= 2, cnt[2 * i]++;
  }
  ans[0] = 1;
  for (int i = 1; i <= n; i++) {
    if (cnt[i] == 1) {
      ans |= (ans << i);
    }
    if (cnt[i] == 2) {
      ans |= (ans << i) | (ans << (2 * i));
    }
  }
  ans1 = k + 1 - ans[k];
  ans2 = 2 * min(sm, k);
  if (sm < k) {
    ans2 += min(k - sm, cn1);
  }
  printf("%d %d\n", ans1, ans2);
  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 = 1e6 + 10, mod = 1e9 + 7;
const long long inf = 1e18;
int f[maxn], cnt[maxn], cn1, sm, ans1, ans2;
bool mark[maxn];
bitset<maxn> ans;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &f[i]);
  for (int i = 1; i <= n; i++) {
    if (!mark[i]) {
      int tmp = i, ln = 0;
      while (!mark[tmp]) {
        mark[tmp] = 1;
        ln++;
        tmp = f[tmp];
      }
      sm += ln / 2;
      cn1 += ln & 1;
      cnt[ln]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    while (cnt[i] > 2) cnt[i] -= 2, cnt[2 * i]++;
  }
  ans[0] = 1;
  for (int i = 1; i <= n; i++) {
    if (cnt[i] == 1) {
      ans |= (ans << i);
    }
    if (cnt[i] == 2) {
      ans |= (ans << i) | (ans << (2 * i));
    }
  }
  ans1 = k + 1 - ans[k];
  ans2 = 2 * min(sm, k);
  if (sm < k) {
    ans2 += min(k - sm, cn1);
  }
  printf("%d %d\n", ans1, ans2);
  return 0;
}

```