#include <bits/stdc++.h>
using namespace std;
const long long N = 2e6 + 30, Mod = 1e9 + 7;
const long long SQ = 330;
long long P[N], M[N], cnt[N];
bitset<N> dp;
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  long long n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> P[i];
  long long t = 0;
  for (int i = 1; i <= n; i++) {
    long long x = i;
    long long cur = 0;
    while (!M[x]) {
      M[x] = 1;
      cur++;
      x = P[x];
    }
    t += (cur & 1);
    cnt[cur]++;
  }
  for (int i = 1; i <= n; i++) {
    while (cnt[i] > 2) {
      cnt[i] -= 2;
      cnt[i << 1]++;
    }
  }
  dp[0] = 1;
  for (int i = 1; i <= n; i++)
    while (cnt[i]--) dp |= dp << i;
  if (dp[k])
    cout << k << ' ';
  else
    cout << k + 1 << ' ';
  long long ans = (n - t) >> 1;
  if (k <= ans)
    cout << 2 * k;
  else
    cout << min(k + ans, n);
  return (0);
}

### Prompt

Develop a solution in cpp to the problem described below:

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 = 2e6 + 30, Mod = 1e9 + 7;
const long long SQ = 330;
long long P[N], M[N], cnt[N];
bitset<N> dp;
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  long long n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> P[i];
  long long t = 0;
  for (int i = 1; i <= n; i++) {
    long long x = i;
    long long cur = 0;
    while (!M[x]) {
      M[x] = 1;
      cur++;
      x = P[x];
    }
    t += (cur & 1);
    cnt[cur]++;
  }
  for (int i = 1; i <= n; i++) {
    while (cnt[i] > 2) {
      cnt[i] -= 2;
      cnt[i << 1]++;
    }
  }
  dp[0] = 1;
  for (int i = 1; i <= n; i++)
    while (cnt[i]--) dp |= dp << i;
  if (dp[k])
    cout << k << ' ';
  else
    cout << k + 1 << ' ';
  long long ans = (n - t) >> 1;
  if (k <= ans)
    cout << 2 * k;
  else
    cout << min(k + ans, n);
  return (0);
}

```