#include <bits/stdc++.h>
using namespace std;
int n;
int k;
int belong[1000002];
int cnt[1000002];
inline int root(int b) {
  if (belong[b] == -1) {
    return b;
  }
  belong[b] = root(belong[b]);
  return belong[b];
}
void merge(int a, int b) {
  a = root(a);
  b = root(b);
  if (a == b) {
    return;
  }
  cnt[b] += cnt[a];
  belong[a] = b;
}
map<int, int> mp;
int use[1000002];
int u_s;
bool dp[1000002];
bitset<1000002> b;
int main() {
  b.set(0, true);
  memset(belong, -1, sizeof(belong));
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cnt[i] = 1;
  }
  for (int i = 0; i < n; i++) {
    int p;
    scanf("%d", &p);
    p--;
    merge(i, p);
  }
  for (int i = 0; i < n; i++) {
    if (root(i) == i) {
      mp[cnt[i]]++;
    }
  }
  for (auto it : mp) {
    int offset = it.first;
    int rem = it.second;
    int p = 1;
    while (rem) {
      int cn = min(p, rem);
      b |= (b << (offset * cn));
      rem -= cn;
      p *= 2;
    }
  }
  if (b.test(k)) {
    cout << k << " ";
  } else {
    cout << k + 1 << " ";
  }
  int odd = 0;
  int ev = 0;
  for (int i = 0; i < n; i++) {
    if (root(i) == i) {
      if (cnt[i] & 1) {
        odd++;
      }
      ev += cnt[i] / 2;
    }
  }
  cout << min(n, min(ev, k) * 2 + max(0, k - ev)) << endl;
  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;
int n;
int k;
int belong[1000002];
int cnt[1000002];
inline int root(int b) {
  if (belong[b] == -1) {
    return b;
  }
  belong[b] = root(belong[b]);
  return belong[b];
}
void merge(int a, int b) {
  a = root(a);
  b = root(b);
  if (a == b) {
    return;
  }
  cnt[b] += cnt[a];
  belong[a] = b;
}
map<int, int> mp;
int use[1000002];
int u_s;
bool dp[1000002];
bitset<1000002> b;
int main() {
  b.set(0, true);
  memset(belong, -1, sizeof(belong));
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cnt[i] = 1;
  }
  for (int i = 0; i < n; i++) {
    int p;
    scanf("%d", &p);
    p--;
    merge(i, p);
  }
  for (int i = 0; i < n; i++) {
    if (root(i) == i) {
      mp[cnt[i]]++;
    }
  }
  for (auto it : mp) {
    int offset = it.first;
    int rem = it.second;
    int p = 1;
    while (rem) {
      int cn = min(p, rem);
      b |= (b << (offset * cn));
      rem -= cn;
      p *= 2;
    }
  }
  if (b.test(k)) {
    cout << k << " ";
  } else {
    cout << k + 1 << " ";
  }
  int odd = 0;
  int ev = 0;
  for (int i = 0; i < n; i++) {
    if (root(i) == i) {
      if (cnt[i] & 1) {
        odd++;
      }
      ev += cnt[i] / 2;
    }
  }
  cout << min(n, min(ev, k) * 2 + max(0, k - ev)) << endl;
  return 0;
}

```