#include <bits/stdc++.h>
using namespace std;
const int NMAX = 1000001;
bool vis[NMAX];
int p[NMAX];
int go(int u) {
  vis[u] = true;
  if (vis[p[u]]) return 1;
  return 1 + go(p[u]);
}
int sizes[NMAX];
int cap[NMAX];
bitset<NMAX> knp;
int main() {
  vector<int> v;
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", p + i);
  for (int i = 1; i <= n; i++)
    if (!vis[i]) v.push_back(go(i));
  int one = 0, two = 0;
  for (int c : v) {
    if (c % 2 == 0) {
      two += c / 2;
    } else {
      one++;
      two += c / 2;
    }
  }
  int s = 0;
  if (one + two <= k)
    s = n;
  else {
    if (k <= two)
      s = 2 * k;
    else
      s = 2 * two + (k - two);
  }
  for (int c : v) sizes[c]++;
  const int T = 200;
  knp[0] = true;
  for (int i = 1; i <= T; i++)
    if (sizes[i]) {
      memset(cap, -1, sizeof(cap));
      for (int j = 0; j <= k; j++)
        if (knp[j]) cap[j] = 0;
      for (int j = i; j <= k; j++)
        if (cap[j] < 0 && cap[j - i] >= 0) cap[j] = cap[j - i] + 1;
      for (int j = i; j <= k; j++)
        if (cap[j] > 0 && cap[j] <= sizes[i]) knp[j] = true;
    }
  for (int i = T + 1; i <= k; i++)
    for (int j = 0; j < sizes[i]; j++) {
      knp |= (knp << i);
    }
  printf("%d %d", knp[k] ? k : k + 1, s);
}

### Prompt

Please formulate 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;
const int NMAX = 1000001;
bool vis[NMAX];
int p[NMAX];
int go(int u) {
  vis[u] = true;
  if (vis[p[u]]) return 1;
  return 1 + go(p[u]);
}
int sizes[NMAX];
int cap[NMAX];
bitset<NMAX> knp;
int main() {
  vector<int> v;
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", p + i);
  for (int i = 1; i <= n; i++)
    if (!vis[i]) v.push_back(go(i));
  int one = 0, two = 0;
  for (int c : v) {
    if (c % 2 == 0) {
      two += c / 2;
    } else {
      one++;
      two += c / 2;
    }
  }
  int s = 0;
  if (one + two <= k)
    s = n;
  else {
    if (k <= two)
      s = 2 * k;
    else
      s = 2 * two + (k - two);
  }
  for (int c : v) sizes[c]++;
  const int T = 200;
  knp[0] = true;
  for (int i = 1; i <= T; i++)
    if (sizes[i]) {
      memset(cap, -1, sizeof(cap));
      for (int j = 0; j <= k; j++)
        if (knp[j]) cap[j] = 0;
      for (int j = i; j <= k; j++)
        if (cap[j] < 0 && cap[j - i] >= 0) cap[j] = cap[j - i] + 1;
      for (int j = i; j <= k; j++)
        if (cap[j] > 0 && cap[j] <= sizes[i]) knp[j] = true;
    }
  for (int i = T + 1; i <= k; i++)
    for (int j = 0; j < sizes[i]; j++) {
      knp |= (knp << i);
    }
  printf("%d %d", knp[k] ? k : k + 1, s);
}

```