#include <bits/stdc++.h>
using namespace std;
const long long N = 1e6 + 100, OO = 1e6 + 1000, M = 1e9 + 7, P = 6151,
                SQ = 1000, lg = 30;
long long p[N], num = 0;
bool mark[N];
bitset<OO> b;
map<long long, long long> mp;
void dfs(long long x) {
  mark[x] = true;
  num++;
  if (mark[p[x]]) return;
  dfs(p[x]);
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) cin >> p[i];
  long long ans = 0, t = 0;
  for (long long i = 1; i <= n; i++) {
    if (!mark[i]) {
      num = 0;
      dfs(i);
      mp[num]++;
      ans += (num / 2);
      if (num % 2 == 1) t++;
    }
  }
  b[0] = 1;
  for (auto u : mp) {
    long long x = u.first, y = u.second;
    if (y == 0) continue;
    mp[x * 2] += ((y - 1) / 2);
    b = (b << x) | b;
    if (y % 2 == 0) b = (b << x) | b;
  }
  if (b[m] == 1)
    cout << m << " ";
  else
    cout << m + 1 << " ";
  if (m <= ans)
    cout << m * 2 << endl;
  else {
    if (m - ans <= t)
      cout << 2 * ans + m - ans << endl;
    else
      cout << n << endl;
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to 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 long long N = 1e6 + 100, OO = 1e6 + 1000, M = 1e9 + 7, P = 6151,
                SQ = 1000, lg = 30;
long long p[N], num = 0;
bool mark[N];
bitset<OO> b;
map<long long, long long> mp;
void dfs(long long x) {
  mark[x] = true;
  num++;
  if (mark[p[x]]) return;
  dfs(p[x]);
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) cin >> p[i];
  long long ans = 0, t = 0;
  for (long long i = 1; i <= n; i++) {
    if (!mark[i]) {
      num = 0;
      dfs(i);
      mp[num]++;
      ans += (num / 2);
      if (num % 2 == 1) t++;
    }
  }
  b[0] = 1;
  for (auto u : mp) {
    long long x = u.first, y = u.second;
    if (y == 0) continue;
    mp[x * 2] += ((y - 1) / 2);
    b = (b << x) | b;
    if (y % 2 == 0) b = (b << x) | b;
  }
  if (b[m] == 1)
    cout << m << " ";
  else
    cout << m + 1 << " ";
  if (m <= ans)
    cout << m * 2 << endl;
  else {
    if (m - ans <= t)
      cout << 2 * ans + m - ans << endl;
    else
      cout << n << endl;
  }
  return 0;
}

```