#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
inline int read() {
  int x = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    ;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return x;
}
bitset<N> B;
int n, p[N], k, K, ans[2], lef, c[N];
bool v[N];
inline int getlen(int x) {
  int an = 0;
  for (; !v[x]; x = p[x]) v[x] = 1, an++;
  return an;
}
inline void solve0() {
  for (int i = 1, now; i <= n; i++)
    if (!v[i]) {
      now = getlen(i), c[now]++;
      lef += now & 1, now ^= now & 1;
      if (K >= (now >> 1))
        K -= now >> 1, ans[0] += now;
      else if (K)
        ans[0] += K << 1, K = 0;
    }
  if (K) ans[0] += min(K, lef);
}
inline void solve1() {
  B[0] = 1;
  for (int i = 1; i <= n; i++)
    if (c[i]) {
      B |= B << i, c[i]--;
      if (c[i] & 1) B |= B << i;
      c[i << 1] += c[i] >> 1;
    }
  ans[1] = B[k] ? k : k + 1;
}
int main() {
  n = read(), K = k = read();
  for (int i = 1; i <= n; i++) p[i] = read();
  solve0(), solve1();
  printf("%d %d\n", ans[1], ans[0]);
  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;
const int N = 1e6 + 5;
inline int read() {
  int x = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    ;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return x;
}
bitset<N> B;
int n, p[N], k, K, ans[2], lef, c[N];
bool v[N];
inline int getlen(int x) {
  int an = 0;
  for (; !v[x]; x = p[x]) v[x] = 1, an++;
  return an;
}
inline void solve0() {
  for (int i = 1, now; i <= n; i++)
    if (!v[i]) {
      now = getlen(i), c[now]++;
      lef += now & 1, now ^= now & 1;
      if (K >= (now >> 1))
        K -= now >> 1, ans[0] += now;
      else if (K)
        ans[0] += K << 1, K = 0;
    }
  if (K) ans[0] += min(K, lef);
}
inline void solve1() {
  B[0] = 1;
  for (int i = 1; i <= n; i++)
    if (c[i]) {
      B |= B << i, c[i]--;
      if (c[i] & 1) B |= B << i;
      c[i << 1] += c[i] >> 1;
    }
  ans[1] = B[k] ? k : k + 1;
}
int main() {
  n = read(), K = k = read();
  for (int i = 1; i <= n; i++) p[i] = read();
  solve0(), solve1();
  printf("%d %d\n", ans[1], ans[0]);
  return 0;
}

```