#include <bits/stdc++.h>
using namespace std;
int read(void) {
  int x = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  return x;
}
const int N = 1000500;
int vis[N], to[N], n;
int k, tot, ansx, ansn;
int num[N], w[N];
bitset<N> dp;
int val[N], tot1;
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; i++) to[i] = read();
  for (int i = 1; i <= n; i++) {
    int index = i, cnt = 0;
    while (!vis[index]) {
      vis[index] = 1;
      index = to[index];
      cnt++;
    }
    if (!cnt) continue;
    if (num[cnt])
      num[cnt]++;
    else
      num[cnt]++, w[++tot] = cnt;
  }
  for (int i = 1; i <= tot; i++) ansx += num[w[i]] * (w[i] / 2);
  if (ansx >= k)
    ansx = 2 * k;
  else
    ansx += k;
  ansx = min(ansx, n);
  for (int i = 1; i <= tot; i++) {
    for (int j = 1; j <= num[w[i]]; j <<= 1) {
      val[++tot1] = j * w[i];
      num[w[i]] -= j;
    }
    if (num[w[i]] > 0) val[++tot1] = num[w[i]] * w[i];
  }
  dp.set(0);
  for (int i = 1; i <= tot1; i++) dp |= (dp << val[i]);
  if (dp.test(k))
    ansn = k;
  else
    ansn = k + 1;
  cout << ansn << ' ' << ansx << endl;
  return 0;
}

### Prompt

Generate 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;
int read(void) {
  int x = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  return x;
}
const int N = 1000500;
int vis[N], to[N], n;
int k, tot, ansx, ansn;
int num[N], w[N];
bitset<N> dp;
int val[N], tot1;
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; i++) to[i] = read();
  for (int i = 1; i <= n; i++) {
    int index = i, cnt = 0;
    while (!vis[index]) {
      vis[index] = 1;
      index = to[index];
      cnt++;
    }
    if (!cnt) continue;
    if (num[cnt])
      num[cnt]++;
    else
      num[cnt]++, w[++tot] = cnt;
  }
  for (int i = 1; i <= tot; i++) ansx += num[w[i]] * (w[i] / 2);
  if (ansx >= k)
    ansx = 2 * k;
  else
    ansx += k;
  ansx = min(ansx, n);
  for (int i = 1; i <= tot; i++) {
    for (int j = 1; j <= num[w[i]]; j <<= 1) {
      val[++tot1] = j * w[i];
      num[w[i]] -= j;
    }
    if (num[w[i]] > 0) val[++tot1] = num[w[i]] * w[i];
  }
  dp.set(0);
  for (int i = 1; i <= tot1; i++) dp |= (dp << val[i]);
  if (dp.test(k))
    ansn = k;
  else
    ansn = k + 1;
  cout << ansn << ' ' << ansx << endl;
  return 0;
}

```