#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int n, k, p[N];
bool used[N];
vector<int> q;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
  }
  for (int i = 1; i <= n; i++) {
    if (!used[i]) {
      int cur = 0, v = i;
      while (!used[v]) {
        cur++;
        used[v] = true;
        v = p[v];
      }
      q.push_back(cur);
    }
  }
  bitset<N> dp;
  sort(q.begin(), q.end());
  dp[0] = 1;
  for (int i = 0; i < q.size(); i++) {
    int cnt = 1;
    while (i + 1 < q.size() && q[i] == q[i + 1]) {
      i++;
      cnt++;
    }
    for (int j = 1; j <= cnt; j *= 2) {
      dp |= dp << (j * q[i]);
      cnt -= j;
    }
    dp |= dp << (cnt * q[i]);
  }
  cout << k + !dp[k] << " ";
  int ans = 0;
  for (auto &it : q) {
    while (it >= 2 && k) {
      k--;
      ans += 2;
      it -= 2;
    }
  }
  for (auto &it : q) {
    while (it >= 1 && k) {
      k--;
      ans += 1;
      it -= 1;
    }
  }
  cout << ans;
}

### 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;
const int N = 1e6 + 5;
int n, k, p[N];
bool used[N];
vector<int> q;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
  }
  for (int i = 1; i <= n; i++) {
    if (!used[i]) {
      int cur = 0, v = i;
      while (!used[v]) {
        cur++;
        used[v] = true;
        v = p[v];
      }
      q.push_back(cur);
    }
  }
  bitset<N> dp;
  sort(q.begin(), q.end());
  dp[0] = 1;
  for (int i = 0; i < q.size(); i++) {
    int cnt = 1;
    while (i + 1 < q.size() && q[i] == q[i + 1]) {
      i++;
      cnt++;
    }
    for (int j = 1; j <= cnt; j *= 2) {
      dp |= dp << (j * q[i]);
      cnt -= j;
    }
    dp |= dp << (cnt * q[i]);
  }
  cout << k + !dp[k] << " ";
  int ans = 0;
  for (auto &it : q) {
    while (it >= 2 && k) {
      k--;
      ans += 2;
      it -= 2;
    }
  }
  for (auto &it : q) {
    while (it >= 1 && k) {
      k--;
      ans += 1;
      it -= 1;
    }
  }
  cout << ans;
}

```