#include <bits/stdc++.h>
using namespace std;
void desperate_optimization(int precision) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.setf(ios::fixed);
  cout.setf(ios::showpoint);
  cout.precision(precision);
}
const int N = 1e6;
int stats[N + 5];
int arr[N + 5];
int ctr, idx;
void dfs(int i) {
  stats[i] = idx;
  ctr++;
  if (stats[arr[i]] == 0) dfs(arr[i]);
}
int occ[N + 5];
bitset<N + 5> dp;
int temp[5 * N + 5];
int sz = 0;
int main() {
  desperate_optimization(10);
  clock_t CLOCK;
  CLOCK = clock();
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  int tots = 0;
  for (int i = 1; i <= n; i++) {
    if (stats[i] == 0) {
      idx++;
      ctr = 0;
      dfs(i);
      tots += ctr / 2;
      occ[ctr]++;
    }
  }
  int maxi = 0;
  int mini = 0;
  if (k <= tots)
    maxi = 2 * k;
  else
    maxi = min(n, 2 * tots + (k - tots));
  for (int i = 1; i <= N; i++) {
    if (occ[i] == 0) continue;
    for (int j = 0; (1 << j) <= occ[i]; j++) {
      int val = (1 << j) * i;
      temp[sz++] = val;
      occ[i] -= (1 << j);
    }
    if (occ[i] != 0) temp[sz++] = occ[i] * i;
  }
  dp.set(0);
  for (int i = 0; i < sz; i++) {
    dp |= (dp << temp[i]);
  }
  if (dp.test(k))
    mini = k;
  else
    mini = k + 1;
  cout << mini << " " << maxi << '\n';
  cerr << "PROCESSED TIME " << (clock() - CLOCK) * 1.0 / (1.0 * CLOCKS_PER_SEC)
       << '\n';
  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;
void desperate_optimization(int precision) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.setf(ios::fixed);
  cout.setf(ios::showpoint);
  cout.precision(precision);
}
const int N = 1e6;
int stats[N + 5];
int arr[N + 5];
int ctr, idx;
void dfs(int i) {
  stats[i] = idx;
  ctr++;
  if (stats[arr[i]] == 0) dfs(arr[i]);
}
int occ[N + 5];
bitset<N + 5> dp;
int temp[5 * N + 5];
int sz = 0;
int main() {
  desperate_optimization(10);
  clock_t CLOCK;
  CLOCK = clock();
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  int tots = 0;
  for (int i = 1; i <= n; i++) {
    if (stats[i] == 0) {
      idx++;
      ctr = 0;
      dfs(i);
      tots += ctr / 2;
      occ[ctr]++;
    }
  }
  int maxi = 0;
  int mini = 0;
  if (k <= tots)
    maxi = 2 * k;
  else
    maxi = min(n, 2 * tots + (k - tots));
  for (int i = 1; i <= N; i++) {
    if (occ[i] == 0) continue;
    for (int j = 0; (1 << j) <= occ[i]; j++) {
      int val = (1 << j) * i;
      temp[sz++] = val;
      occ[i] -= (1 << j);
    }
    if (occ[i] != 0) temp[sz++] = occ[i] * i;
  }
  dp.set(0);
  for (int i = 0; i < sz; i++) {
    dp |= (dp << temp[i]);
  }
  if (dp.test(k))
    mini = k;
  else
    mini = k + 1;
  cout << mini << " " << maxi << '\n';
  cerr << "PROCESSED TIME " << (clock() - CLOCK) * 1.0 / (1.0 * CLOCKS_PER_SEC)
       << '\n';
  return 0;
}

```