#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int n, k, p[N];
int cnt[N];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", p + i);
  bitset<N> vis;
  vector<int> v;
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    int cur = i, cnt = 0;
    while (!vis[cur]) {
      cnt++;
      vis[cur] = 1;
      cur = p[cur];
    }
    v.push_back(cnt);
  }
  sort(v.begin(), v.end());
  int tmp = k;
  int ans = 0;
  for (int i = 0; i < v.size(); i++) {
    cnt[v[i]]++;
    int x = min(tmp, v[i] / 2);
    ans += x * 2;
    v[i] -= x * 2;
    tmp -= x;
  }
  for (int i = 0; i < v.size(); i++) {
    int x = min(tmp, v[i]);
    ans += x;
    tmp -= x;
  }
  vis.reset();
  vis[0] = 1;
  int ans2 = k + 1;
  for (int i = 1; i <= k; i++) {
    if (i > 50) {
      for (int j = 0; j < cnt[i]; j++) {
        vis = (vis | (vis << i));
      }
    } else {
      if (!cnt[i]) continue;
      vector<deque<int>> dq(i);
      vector<int> sum(i);
      int cur = 0;
      for (int j = 0; j <= k; j++) {
        dq[cur].push_back(vis[j]);
        sum[cur] += vis[j];
        if (sum[cur]) vis[j] = 1;
        if (dq[cur].size() > cnt[i]) {
          sum[cur] -= dq[j % i].front();
          dq[cur].pop_front();
        }
        ++cur;
        if (cur == i) cur = 0;
      }
    }
  }
  if (vis[k]) ans2 = k;
  cout << ans2 << ' ' << ans << '\n';
  return 0;
}

### Prompt

Create a solution in CPP for 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;
int n, k, p[N];
int cnt[N];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", p + i);
  bitset<N> vis;
  vector<int> v;
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    int cur = i, cnt = 0;
    while (!vis[cur]) {
      cnt++;
      vis[cur] = 1;
      cur = p[cur];
    }
    v.push_back(cnt);
  }
  sort(v.begin(), v.end());
  int tmp = k;
  int ans = 0;
  for (int i = 0; i < v.size(); i++) {
    cnt[v[i]]++;
    int x = min(tmp, v[i] / 2);
    ans += x * 2;
    v[i] -= x * 2;
    tmp -= x;
  }
  for (int i = 0; i < v.size(); i++) {
    int x = min(tmp, v[i]);
    ans += x;
    tmp -= x;
  }
  vis.reset();
  vis[0] = 1;
  int ans2 = k + 1;
  for (int i = 1; i <= k; i++) {
    if (i > 50) {
      for (int j = 0; j < cnt[i]; j++) {
        vis = (vis | (vis << i));
      }
    } else {
      if (!cnt[i]) continue;
      vector<deque<int>> dq(i);
      vector<int> sum(i);
      int cur = 0;
      for (int j = 0; j <= k; j++) {
        dq[cur].push_back(vis[j]);
        sum[cur] += vis[j];
        if (sum[cur]) vis[j] = 1;
        if (dq[cur].size() > cnt[i]) {
          sum[cur] -= dq[j % i].front();
          dq[cur].pop_front();
        }
        ++cur;
        if (cur == i) cur = 0;
      }
    }
  }
  if (vis[k]) ans2 = k;
  cout << ans2 << ' ' << ans << '\n';
  return 0;
}

```