#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000005;
const int MOD = 1000000007;
int u[MAXN], sz[MAXN];
int arr[MAXN], p, cnt[MAXN];
bitset<MAXN> b;
int root(int a) { return u[a] == a ? a : u[a] = root(u[a]); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) u[i] = i, sz[i] = 1;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    if (root(a) != root(i)) {
      sz[root(i)] += sz[root(a)];
      u[root(a)] = root(i);
    }
  }
  int c1 = 0, c2 = 0;
  for (int i = 1; i <= n; i++)
    if (root(i) == i) {
      arr[p++] = sz[i];
      cnt[sz[i]]++;
      c1 += sz[i] / 2;
      c2 += sz[i] % 2;
    }
  b.set(0);
  for (int i = 1; i <= n; i++)
    if (cnt[i]) {
      for (int j = 1; j <= cnt[i]; j *= 2) {
        b |= (b << (i * j));
        cnt[i] -= j;
      }
      b |= (b << (i * cnt[i]));
    }
  int mx = 0, mn = 0;
  mn = k;
  if (!b.test(k)) mn++;
  mx = min(2 * k, 2 * c1);
  mx += min(k - mx / 2, c2);
  cout << mn << ' ' << mx << '\n';
  return 0;
}

### Prompt

Please formulate 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 int MAXN = 1000005;
const int MOD = 1000000007;
int u[MAXN], sz[MAXN];
int arr[MAXN], p, cnt[MAXN];
bitset<MAXN> b;
int root(int a) { return u[a] == a ? a : u[a] = root(u[a]); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) u[i] = i, sz[i] = 1;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    if (root(a) != root(i)) {
      sz[root(i)] += sz[root(a)];
      u[root(a)] = root(i);
    }
  }
  int c1 = 0, c2 = 0;
  for (int i = 1; i <= n; i++)
    if (root(i) == i) {
      arr[p++] = sz[i];
      cnt[sz[i]]++;
      c1 += sz[i] / 2;
      c2 += sz[i] % 2;
    }
  b.set(0);
  for (int i = 1; i <= n; i++)
    if (cnt[i]) {
      for (int j = 1; j <= cnt[i]; j *= 2) {
        b |= (b << (i * j));
        cnt[i] -= j;
      }
      b |= (b << (i * cnt[i]));
    }
  int mx = 0, mn = 0;
  mn = k;
  if (!b.test(k)) mn++;
  mx = min(2 * k, 2 * c1);
  mx += min(k - mx / 2, c2);
  cout << mn << ' ' << mx << '\n';
  return 0;
}

```