#include <bits/stdc++.h>
using namespace std;
const int N = 1000 * 1000 + 10;
int n, k, Odd;
int P[N], cycle[N];
vector<int> gp;
bitset<N> bt;
void input() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> P[i];
  return;
}
void PrePros() {
  for (int i = 1; i <= n; i++) {
    int cnt = 0, pnt = i;
    while (P[pnt] != -1) {
      int last = pnt;
      pnt = P[last];
      P[last] = -1;
      cnt++;
    }
    if (cnt % 2) Odd++;
    cycle[cnt]++;
  }
  for (int i = 1; i <= n / 2; i++)
    if (cycle[i]) {
      cycle[2 * i] += (cycle[i] - 1) / 2;
      cycle[i] = 2 - (cycle[i] % 2);
    }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < cycle[i]; j++) gp.push_back(i);
  return;
}
void solve() {
  PrePros();
  bt[0] = 1;
  while (!gp.empty()) {
    bt |= (bt << gp.back());
    gp.pop_back();
  }
  return;
}
void output() {
  cout << k + !(bt[k]) << " ";
  cout << min(min(2 * k, n - Odd) + max(k - ((n - Odd) / 2), 0), n) << endl;
  return;
}
int main() {
  input();
  solve();
  output();
  return 0;
}

### Prompt

In cpp, your task is to solve 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 N = 1000 * 1000 + 10;
int n, k, Odd;
int P[N], cycle[N];
vector<int> gp;
bitset<N> bt;
void input() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> P[i];
  return;
}
void PrePros() {
  for (int i = 1; i <= n; i++) {
    int cnt = 0, pnt = i;
    while (P[pnt] != -1) {
      int last = pnt;
      pnt = P[last];
      P[last] = -1;
      cnt++;
    }
    if (cnt % 2) Odd++;
    cycle[cnt]++;
  }
  for (int i = 1; i <= n / 2; i++)
    if (cycle[i]) {
      cycle[2 * i] += (cycle[i] - 1) / 2;
      cycle[i] = 2 - (cycle[i] % 2);
    }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < cycle[i]; j++) gp.push_back(i);
  return;
}
void solve() {
  PrePros();
  bt[0] = 1;
  while (!gp.empty()) {
    bt |= (bt << gp.back());
    gp.pop_back();
  }
  return;
}
void output() {
  cout << k + !(bt[k]) << " ";
  cout << min(min(2 * k, n - Odd) + max(k - ((n - Odd) / 2), 0), n) << endl;
  return;
}
int main() {
  input();
  solve();
  output();
  return 0;
}

```