#include <bits/stdc++.h>
using namespace std;
const int maxN = 1e6 + 5;
int P[maxN], C[maxN];
bitset<maxN> S;
bool V[maxN];
vector<int> L;
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> P[i];
  for (int i = 1; i <= n; ++i) {
    if (V[i]) continue;
    V[i] = true;
    int length = 1;
    int j = P[i];
    while (!V[j]) {
      V[j] = true;
      length++;
      j = P[j];
    }
    L.push_back(length);
    C[length]++;
  }
  int max_val = 0, rem = k, not_full = 0;
  for (int i = 0; i < L.size(); ++i) {
    max_val += min((L[i] / 2) * 2, rem * 2);
    rem -= min(L[i] / 2, rem);
    not_full += L[i] % 2;
  }
  max_val += min(rem, not_full);
  S.set(0);
  for (int i = 1; i <= k; ++i) {
    if (C[i] == 0) continue;
    int j = 1;
    while (C[i] != 0) {
      j = min(j, C[i]);
      int m = i * j;
      S |= (S << m);
      C[i] -= j;
      j *= 2;
    }
  }
  int min_val = k + 1;
  if (S.test(k)) min_val--;
  cout << min_val << " " << max_val << "\n";
}

### Prompt

Develop a solution in cpp to the problem described below:

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 maxN = 1e6 + 5;
int P[maxN], C[maxN];
bitset<maxN> S;
bool V[maxN];
vector<int> L;
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> P[i];
  for (int i = 1; i <= n; ++i) {
    if (V[i]) continue;
    V[i] = true;
    int length = 1;
    int j = P[i];
    while (!V[j]) {
      V[j] = true;
      length++;
      j = P[j];
    }
    L.push_back(length);
    C[length]++;
  }
  int max_val = 0, rem = k, not_full = 0;
  for (int i = 0; i < L.size(); ++i) {
    max_val += min((L[i] / 2) * 2, rem * 2);
    rem -= min(L[i] / 2, rem);
    not_full += L[i] % 2;
  }
  max_val += min(rem, not_full);
  S.set(0);
  for (int i = 1; i <= k; ++i) {
    if (C[i] == 0) continue;
    int j = 1;
    while (C[i] != 0) {
      j = min(j, C[i]);
      int m = i * j;
      S |= (S << m);
      C[i] -= j;
      j *= 2;
    }
  }
  int min_val = k + 1;
  if (S.test(k)) min_val--;
  cout << min_val << " " << max_val << "\n";
}

```