#include <bits/stdc++.h>
using namespace std;
const int maxn = 1123456;
bitset<maxn> dp;
bool vis[maxn];
int p[maxn], w[maxn], have[maxn], val[maxn];
int n, k, tot, tt, mmax;
int main() {
  scanf("%d%d", &n, &k);
  for (int(i) = (1); (i) <= (n); (i)++) scanf("%d", p + i);
  for (int(i) = (1); (i) <= (n); (i)++) {
    int now = i, cnt = 0;
    while (!vis[now]) {
      vis[now] = true;
      now = p[now];
      cnt++;
    }
    if (cnt) {
      if (!have[cnt]) w[++tot] = cnt;
      have[cnt]++;
    }
  }
  for (int(i) = (1); (i) <= (tot); (i)++) mmax += w[i] / 2 * have[w[i]];
  if (mmax >= k)
    mmax = 2 * k;
  else
    mmax += k;
  mmax = min(mmax, n);
  for (int(i) = (1); (i) <= (tot); (i)++) {
    for (int j = 1; j <= have[w[i]]; j <<= 1) {
      val[++tt] = j * w[i];
      have[w[i]] -= j;
    }
    if (have[w[i]] > 0) val[++tt] = have[w[i]] * w[i];
  }
  dp.set(0);
  for (int(i) = (1); (i) <= (tt); (i)++) dp |= dp << val[i];
  if (dp[k])
    printf("%d %d", k, mmax);
  else
    printf("%d %d", k + 1, mmax);
}

### Prompt

Create a solution in cpp for 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 = 1123456;
bitset<maxn> dp;
bool vis[maxn];
int p[maxn], w[maxn], have[maxn], val[maxn];
int n, k, tot, tt, mmax;
int main() {
  scanf("%d%d", &n, &k);
  for (int(i) = (1); (i) <= (n); (i)++) scanf("%d", p + i);
  for (int(i) = (1); (i) <= (n); (i)++) {
    int now = i, cnt = 0;
    while (!vis[now]) {
      vis[now] = true;
      now = p[now];
      cnt++;
    }
    if (cnt) {
      if (!have[cnt]) w[++tot] = cnt;
      have[cnt]++;
    }
  }
  for (int(i) = (1); (i) <= (tot); (i)++) mmax += w[i] / 2 * have[w[i]];
  if (mmax >= k)
    mmax = 2 * k;
  else
    mmax += k;
  mmax = min(mmax, n);
  for (int(i) = (1); (i) <= (tot); (i)++) {
    for (int j = 1; j <= have[w[i]]; j <<= 1) {
      val[++tt] = j * w[i];
      have[w[i]] -= j;
    }
    if (have[w[i]] > 0) val[++tt] = have[w[i]] * w[i];
  }
  dp.set(0);
  for (int(i) = (1); (i) <= (tt); (i)++) dp |= dp << val[i];
  if (dp[k])
    printf("%d %d", k, mmax);
  else
    printf("%d %d", k + 1, mmax);
}

```