#include <bits/stdc++.h>
using namespace std;
int n, k;
int p[1000005];
int vis[1000005];
vector<int> cyc;
int cnt[1000005];
int t;
void dfs(int cn) {
  if (vis[cn]) {
    cyc.push_back(t - vis[cn]);
    ++cnt[t - vis[cn]];
    return;
  }
  vis[cn] = t++;
  dfs(p[cn]);
}
bitset<1000005> bleh;
int cnts[1000005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = (1); i <= (signed)(n); i++) {
    cin >> p[i];
  }
  t = 1;
  for (int i = (1); i <= (signed)(n); i++) {
    if (vis[i] == 0) dfs(i);
  }
  sort(cyc.begin(), cyc.end());
  int bd = 0;
  for (auto u : cyc) {
    bd += u / 2;
  }
  bd = min(bd, k);
  int mx = min(n, bd + k);
  bleh[0] = true;
  for (int i = (2); i <= (signed)(n); i++) {
    if (cnt[i] == 0) continue;
    if (i < 200) {
      memset(cnts, 0, sizeof cnts);
      for (int j = (i); j <= (signed)(n); j++) {
        if (!bleh[j] && bleh[j - i] && cnts[j - i] < cnt[i]) {
          bleh[j] = true;
          cnts[j] = cnts[j - i] + 1;
        }
      }
    } else {
      for (int _ = 0; _ < (signed)(cnt[i]); _++) {
        bleh |= (bleh << i);
      }
    }
  }
  int mn;
  if (bleh[k])
    mn = k;
  else
    mn = k + 1;
  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;
int n, k;
int p[1000005];
int vis[1000005];
vector<int> cyc;
int cnt[1000005];
int t;
void dfs(int cn) {
  if (vis[cn]) {
    cyc.push_back(t - vis[cn]);
    ++cnt[t - vis[cn]];
    return;
  }
  vis[cn] = t++;
  dfs(p[cn]);
}
bitset<1000005> bleh;
int cnts[1000005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = (1); i <= (signed)(n); i++) {
    cin >> p[i];
  }
  t = 1;
  for (int i = (1); i <= (signed)(n); i++) {
    if (vis[i] == 0) dfs(i);
  }
  sort(cyc.begin(), cyc.end());
  int bd = 0;
  for (auto u : cyc) {
    bd += u / 2;
  }
  bd = min(bd, k);
  int mx = min(n, bd + k);
  bleh[0] = true;
  for (int i = (2); i <= (signed)(n); i++) {
    if (cnt[i] == 0) continue;
    if (i < 200) {
      memset(cnts, 0, sizeof cnts);
      for (int j = (i); j <= (signed)(n); j++) {
        if (!bleh[j] && bleh[j - i] && cnts[j - i] < cnt[i]) {
          bleh[j] = true;
          cnts[j] = cnts[j - i] + 1;
        }
      }
    } else {
      for (int _ = 0; _ < (signed)(cnt[i]); _++) {
        bleh |= (bleh << i);
      }
    }
  }
  int mn;
  if (bleh[k])
    mn = k;
  else
    mn = k + 1;
  cout << mn << " " << mx << "\n";
  return 0;
}

```