#include <bits/stdc++.h>
using namespace std;
constexpr int N = 1e6 + 10;
constexpr int LG = 20;
constexpr int MOD = 1e9 + 7;
constexpr int MOD2 = 1e9 + 9;
int n, k, p[N], sz, cnt[N], mx;
bitset<N> foo;
bool mark[N];
void dfs(int v) {
  mark[v] = true;
  sz++;
  if (!mark[p[v]]) dfs(p[v]);
}
int main() {
  ios_base::sync_with_stdio(false);
  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 (!mark[i]) {
      sz = 0;
      dfs(i);
      cnt[sz]++;
    }
  }
  int lim = k;
  for (int i = 1; i <= n; i++) {
    if (lim == 0) break;
    for (int j = 0; j < cnt[i]; j++) {
      if (lim >= i / 2) {
        lim -= i / 2;
        mx += 2 * (i / 2);
      } else {
        mx += 2 * lim;
        lim = 0;
      }
    }
  }
  for (int i = 1; i <= n; i += 2) {
    if (lim <= cnt[i]) {
      mx += lim;
      lim = 0;
      break;
    } else {
      lim -= cnt[i];
      mx += cnt[i];
    }
  }
  foo.set(0);
  for (int i = 1; i <= n; i++) {
    if (cnt[i]) {
      int p = 1;
      while (p <= cnt[i]) {
        cnt[i] -= p;
        foo |= foo << (p * i);
        p <<= 1;
      }
      foo |= foo << (cnt[i] * i);
    }
  }
  if (foo[k])
    cout << k << ' ';
  else
    cout << k + 1 << ' ';
  cout << mx << '\n';
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
constexpr int N = 1e6 + 10;
constexpr int LG = 20;
constexpr int MOD = 1e9 + 7;
constexpr int MOD2 = 1e9 + 9;
int n, k, p[N], sz, cnt[N], mx;
bitset<N> foo;
bool mark[N];
void dfs(int v) {
  mark[v] = true;
  sz++;
  if (!mark[p[v]]) dfs(p[v]);
}
int main() {
  ios_base::sync_with_stdio(false);
  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 (!mark[i]) {
      sz = 0;
      dfs(i);
      cnt[sz]++;
    }
  }
  int lim = k;
  for (int i = 1; i <= n; i++) {
    if (lim == 0) break;
    for (int j = 0; j < cnt[i]; j++) {
      if (lim >= i / 2) {
        lim -= i / 2;
        mx += 2 * (i / 2);
      } else {
        mx += 2 * lim;
        lim = 0;
      }
    }
  }
  for (int i = 1; i <= n; i += 2) {
    if (lim <= cnt[i]) {
      mx += lim;
      lim = 0;
      break;
    } else {
      lim -= cnt[i];
      mx += cnt[i];
    }
  }
  foo.set(0);
  for (int i = 1; i <= n; i++) {
    if (cnt[i]) {
      int p = 1;
      while (p <= cnt[i]) {
        cnt[i] -= p;
        foo |= foo << (p * i);
        p <<= 1;
      }
      foo |= foo << (cnt[i] * i);
    }
  }
  if (foo[k])
    cout << k << ' ';
  else
    cout << k + 1 << ' ';
  cout << mx << '\n';
  return 0;
}

```