PolandBall and Gifts

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, p _i ≠ 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 p _x = 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 p _i is the index of Ball who should get a gift from the i-th Ball. For all i, p _i ≠ 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.

Solution:

#include <bits/stdc++.h>

using namespace std;

const int N = 1000010;

bitset<N> b;
int p[N];
int cnt[N];
bool was[N];

int main() {
int n, k;
scanf("%d %d", &n, &k);
for (int i = 0; i < n; i++) {
    scanf("%d", p + i);
    p[i]--;
    was[i] = false;
  }
  for (int i = 1; i <= n; i++) {
    cnt[i] = 0;
  }
  int c1 = 0, c2 = 0;
  for (int i = 0; i < n; i++) {
    if (was[i]) {
      continue;
    }
    int x = i;
    int len = 0;
    while (!was[x]) {
      was[x] = true;
      x = p[x];
      len++;
    }
    c1 += len % 2;
    c2 += len / 2;
    cnt[len]++;
  }
  b.set(n);
  for (int i = 1; i <= n; i++) {
    if (cnt[i] == 0) {
      continue;
    }
    int j = 1;
    while (cnt[i] > 0) {
j = min(j, cnt[i]);
int u = i * j;
b |= (b >> u);
cnt[i] -= j;
j *= 2;
}
}
int ans_min = k;
if (!b.test(n - k)) {
ans_min++;
}
int ans_max = 0;
ans_max += 2 * min(k, c2);
k -= min(k, c2);
ans_max += min(k, c1);
printf("%d %d\n", ans_min, ans_max);
return 0;
}