#include <bits/stdc++.h>
using namespace std;
const long double PI = acos(-1.0);
const int MAXN = 2000001;
int a[MAXN], ac = 0;
bool w[MAXN];
int p[MAXN];
int cn[MAXN];
int f[MAXN];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  memset(w, 0, sizeof(w));
  memset(cn, 0, sizeof(cn));
  for (int i = 1; i <= n; i++) {
    scanf("%d", p + i);
  }
  int c2 = 0;
  for (int i = 1; i <= n; i++) {
    if (w[i]) continue;
    int ci = i, cc = 0;
    while (!w[ci]) {
      cc++;
      w[ci] = true;
      ci = p[ci];
    }
    a[ac++] = cc;
    cn[cc]++;
    c2 += cc / 2;
  }
  int sm = 0;
  for (int i = 1; i <= n; i++) f[i] = MAXN;
  f[0] = 0;
  bool og = true;
  for (int i = 1; i <= n; i++) {
    int ad = i * cn[i];
    if (cn[i] == 0) continue;
    sm += ad;
    for (int j = 0; j + i <= sm; j++) {
      int nv = f[j] + 1;
      if (f[j + i] > nv) f[j + i] = nv;
    }
    for (int j = 0; j <= sm; j++) {
      if (f[j] <= cn[i])
        f[j] = 0;
      else {
        f[j] = MAXN;
      }
    }
  }
  if (f[k] == 0)
    cout << k << " ";
  else
    cout << k + 1 << " ";
  cout << min(n, k + min(k, c2)) << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 long double PI = acos(-1.0);
const int MAXN = 2000001;
int a[MAXN], ac = 0;
bool w[MAXN];
int p[MAXN];
int cn[MAXN];
int f[MAXN];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  memset(w, 0, sizeof(w));
  memset(cn, 0, sizeof(cn));
  for (int i = 1; i <= n; i++) {
    scanf("%d", p + i);
  }
  int c2 = 0;
  for (int i = 1; i <= n; i++) {
    if (w[i]) continue;
    int ci = i, cc = 0;
    while (!w[ci]) {
      cc++;
      w[ci] = true;
      ci = p[ci];
    }
    a[ac++] = cc;
    cn[cc]++;
    c2 += cc / 2;
  }
  int sm = 0;
  for (int i = 1; i <= n; i++) f[i] = MAXN;
  f[0] = 0;
  bool og = true;
  for (int i = 1; i <= n; i++) {
    int ad = i * cn[i];
    if (cn[i] == 0) continue;
    sm += ad;
    for (int j = 0; j + i <= sm; j++) {
      int nv = f[j] + 1;
      if (f[j + i] > nv) f[j + i] = nv;
    }
    for (int j = 0; j <= sm; j++) {
      if (f[j] <= cn[i])
        f[j] = 0;
      else {
        f[j] = MAXN;
      }
    }
  }
  if (f[k] == 0)
    cout << k << " ";
  else
    cout << k + 1 << " ";
  cout << min(n, k + min(k, c2)) << endl;
  return 0;
}

```