#include <bits/stdc++.h>
using namespace std;
typedef const int &ci;
const int N = 1000006;
inline int read() {
  int x = 0;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while ('0' <= c && c <= '9') x = 10 * x + c - '0', c = getchar();
  return x;
}
bool vis[N];
int n, k, p[N];
int cnt[N];
int f[N], ans;
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; i++) p[i] = read();
  for (int i = 1, u, tot; i <= n; i++)
    if (!vis[i]) {
      u = i, tot = 0;
      while (!vis[u]) {
        vis[u] = true;
        tot++, u = p[u];
      }
      ans += (tot >> 1);
      cnt[tot]++;
    }
  f[0] = true;
  for (int w = 1, pos, mx = 0; w <= n; w++)
    if (cnt[w]) {
      mx += w * cnt[w];
      for (int r = 0; r < w; r++) {
        pos = -0x3f3f3f3f;
        for (int i = r; i <= mx; i += w) {
          if (f[i]) pos = i;
          if (i - pos <= cnt[w] * w) f[i] = true;
        }
      }
    }
  printf("%d %d\n", k + 1 - f[k], min(n, k > ans ? ans + k : 2 * k));
  return 0;
}

### 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;
typedef const int &ci;
const int N = 1000006;
inline int read() {
  int x = 0;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while ('0' <= c && c <= '9') x = 10 * x + c - '0', c = getchar();
  return x;
}
bool vis[N];
int n, k, p[N];
int cnt[N];
int f[N], ans;
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; i++) p[i] = read();
  for (int i = 1, u, tot; i <= n; i++)
    if (!vis[i]) {
      u = i, tot = 0;
      while (!vis[u]) {
        vis[u] = true;
        tot++, u = p[u];
      }
      ans += (tot >> 1);
      cnt[tot]++;
    }
  f[0] = true;
  for (int w = 1, pos, mx = 0; w <= n; w++)
    if (cnt[w]) {
      mx += w * cnt[w];
      for (int r = 0; r < w; r++) {
        pos = -0x3f3f3f3f;
        for (int i = r; i <= mx; i += w) {
          if (f[i]) pos = i;
          if (i - pos <= cnt[w] * w) f[i] = true;
        }
      }
    }
  printf("%d %d\n", k + 1 - f[k], min(n, k > ans ? ans + k : 2 * k));
  return 0;
}

```