#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5, SQ = 100;
int n, k, p[N], cnt[SQ];
bool mark[N];
vector<int> w, c;
bitset<N> dp1;
int dp2[2][N];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
  }
  for (int i = 0; i < n; i++) {
    if (!mark[i]) {
      int v = i, cnt = 0;
      while (!mark[v]) {
        mark[v] = true;
        cnt++;
        v = p[v];
      }
      c.push_back(cnt);
    }
  }
  int m = c.size();
  int mx = 0, cur = 0, rem = 0;
  for (int i = 0; i < m; i++) cur += c[i] / 2, rem += c[i] & 1;
  if (cur >= k)
    mx = 2 * k;
  else
    mx = cur * 2 + min(k - cur, rem);
  for (int i = 0; i < m; i++) {
    if (c[i] >= SQ)
      w.push_back(c[i]);
    else
      cnt[c[i]]++;
  }
  dp1[0] = true;
  for (int i = 0; i < w.size(); i++) dp1 |= (dp1 << w[i]);
  memset(dp2, -1, sizeof dp2);
  dp2[0][0] = 0;
  for (int i = 0; i < SQ; i++) {
    for (int j = 0; j <= k; j++) {
      int x = (i + 1) & 1;
      if (dp2[x ^ 1][j] >= 0)
        dp2[x][j] = 0;
      else if (j >= i && dp2[x][j - i] >= 0 && dp2[x][j - i] < cnt[i])
        dp2[x][j] = dp2[x][j - i] + 1;
      else
        dp2[x][j] = -1;
    }
  }
  int mn = k + 1;
  for (int i = 0; i <= k; i++)
    if (dp1[i] == true && dp2[SQ & 1][k - i] != -1) mn = k;
  cout << mn << " " << mx << "\n";
}

### Prompt

Generate 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 N = 1e6 + 5, SQ = 100;
int n, k, p[N], cnt[SQ];
bool mark[N];
vector<int> w, c;
bitset<N> dp1;
int dp2[2][N];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
  }
  for (int i = 0; i < n; i++) {
    if (!mark[i]) {
      int v = i, cnt = 0;
      while (!mark[v]) {
        mark[v] = true;
        cnt++;
        v = p[v];
      }
      c.push_back(cnt);
    }
  }
  int m = c.size();
  int mx = 0, cur = 0, rem = 0;
  for (int i = 0; i < m; i++) cur += c[i] / 2, rem += c[i] & 1;
  if (cur >= k)
    mx = 2 * k;
  else
    mx = cur * 2 + min(k - cur, rem);
  for (int i = 0; i < m; i++) {
    if (c[i] >= SQ)
      w.push_back(c[i]);
    else
      cnt[c[i]]++;
  }
  dp1[0] = true;
  for (int i = 0; i < w.size(); i++) dp1 |= (dp1 << w[i]);
  memset(dp2, -1, sizeof dp2);
  dp2[0][0] = 0;
  for (int i = 0; i < SQ; i++) {
    for (int j = 0; j <= k; j++) {
      int x = (i + 1) & 1;
      if (dp2[x ^ 1][j] >= 0)
        dp2[x][j] = 0;
      else if (j >= i && dp2[x][j - i] >= 0 && dp2[x][j - i] < cnt[i])
        dp2[x][j] = dp2[x][j - i] + 1;
      else
        dp2[x][j] = -1;
    }
  }
  int mn = k + 1;
  for (int i = 0; i <= k; i++)
    if (dp1[i] == true && dp2[SQ & 1][k - i] != -1) mn = k;
  cout << mn << " " << mx << "\n";
}

```