#include <bits/stdc++.h>
using namespace std;
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + ch - '0', ch = getchar();
  return x * f;
}
const int inf = 0x3f3f3f3f;
const int N = 1000005;
int n, k, m, sum, tot, cnt, vis[N], cir[N], dp[N], a[N], Max, Min, len, v[N], t;
int main() {
  n = read();
  k = read();
  sum = n;
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++)
    if (!vis[i]) {
      cnt = 1;
      for (int j = a[i]; j != i; j = a[j]) cnt++, vis[j] = 1;
      cir[++tot] = cnt;
      sum -= (cnt & 1);
    }
  sort(cir + 1, cir + tot + 1);
  if (k * 2 <= sum)
    Max = k * 2;
  else
    Max = min(sum + k - sum / 2, n);
  for (int i = 1, j; i <= tot; i++)
    if (cir[i] != cir[i + 1]) {
      len++;
      for (j = 1; len; j <<= 1) t = min(len, j), v[++m] = t * cir[i], len -= t;
    } else
      len++;
  dp[0] = 1;
  for (int i = 1; i <= m; i++)
    for (int j = k; j >= v[i]; j--) dp[j] = dp[j] | dp[j - v[i]];
  Min = k + (!dp[k]);
  printf("%d %d\n", Min, Max);
  return 0;
}

### Prompt

Your task is to create 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() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + ch - '0', ch = getchar();
  return x * f;
}
const int inf = 0x3f3f3f3f;
const int N = 1000005;
int n, k, m, sum, tot, cnt, vis[N], cir[N], dp[N], a[N], Max, Min, len, v[N], t;
int main() {
  n = read();
  k = read();
  sum = n;
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++)
    if (!vis[i]) {
      cnt = 1;
      for (int j = a[i]; j != i; j = a[j]) cnt++, vis[j] = 1;
      cir[++tot] = cnt;
      sum -= (cnt & 1);
    }
  sort(cir + 1, cir + tot + 1);
  if (k * 2 <= sum)
    Max = k * 2;
  else
    Max = min(sum + k - sum / 2, n);
  for (int i = 1, j; i <= tot; i++)
    if (cir[i] != cir[i + 1]) {
      len++;
      for (j = 1; len; j <<= 1) t = min(len, j), v[++m] = t * cir[i], len -= t;
    } else
      len++;
  dp[0] = 1;
  for (int i = 1; i <= m; i++)
    for (int j = k; j >= v[i]; j--) dp[j] = dp[j] | dp[j - v[i]];
  Min = k + (!dp[k]);
  printf("%d %d\n", Min, Max);
  return 0;
}

```