#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 1;
const int INF = 1e9 + 10;
const long long MOD = 1e9 + 7;
int a[N];
bool used[N];
const int T = 100;
bitset<N> bs;
int cnt_used_x[N];
void solve1(vector<int>& lens, int n, int k) {
  sort(lens.rbegin(), lens.rend());
  bs[0] = 1;
  for (auto x : lens) {
    if (x < T) break;
    bs |= (bs << x);
  }
  unordered_map<int, int> cnt_by_x;
  for (auto x : lens) {
    if (x < T) {
      cnt_by_x[x]++;
    }
  }
  for (auto [x, cnt] : cnt_by_x) {
    fill(cnt_used_x, cnt_used_x + k + 1, 0);
    for (int i = 0; i <= k; ++i) {
      if (!bs[i] || cnt_used_x[i] >= cnt) continue;
      if (i + x <= k) {
        if (bs[i + x]) {
          cnt_used_x[i + x] = min(cnt_used_x[i + x], cnt_used_x[i] + 1);
        } else {
          bs[i + x] = 1;
          cnt_used_x[i + x] = cnt_used_x[i] + 1;
        }
      }
    }
  }
  if (bs[k]) {
    cout << k << " ";
  } else {
    cout << k + 1 << " ";
  }
}
void solve() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    a[i]--;
  }
  vector<int> lens;
  for (int i = 0; i < n; ++i) {
    if (!used[i]) {
      lens.push_back(1);
      int cur = a[i];
      used[i] = true;
      while (cur != i) {
        used[cur] = true;
        lens.back()++;
        cur = a[cur];
      }
    }
  }
  int cnt_even = 0, cnt_odd = 0;
  for (auto l : lens) {
    if (l % 2) {
      cnt_odd++;
    } else {
      cnt_even++;
    }
  }
  solve1(lens, n, k);
  int k1 = k;
  int ans2 = min(k1, (n - cnt_odd) / 2) * 2;
  k1 -= min(k1, (n - cnt_odd) / 2);
  ans2 += min(k1, cnt_odd);
  cout << ans2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
}

### 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 N = 1e6 + 1;
const int INF = 1e9 + 10;
const long long MOD = 1e9 + 7;
int a[N];
bool used[N];
const int T = 100;
bitset<N> bs;
int cnt_used_x[N];
void solve1(vector<int>& lens, int n, int k) {
  sort(lens.rbegin(), lens.rend());
  bs[0] = 1;
  for (auto x : lens) {
    if (x < T) break;
    bs |= (bs << x);
  }
  unordered_map<int, int> cnt_by_x;
  for (auto x : lens) {
    if (x < T) {
      cnt_by_x[x]++;
    }
  }
  for (auto [x, cnt] : cnt_by_x) {
    fill(cnt_used_x, cnt_used_x + k + 1, 0);
    for (int i = 0; i <= k; ++i) {
      if (!bs[i] || cnt_used_x[i] >= cnt) continue;
      if (i + x <= k) {
        if (bs[i + x]) {
          cnt_used_x[i + x] = min(cnt_used_x[i + x], cnt_used_x[i] + 1);
        } else {
          bs[i + x] = 1;
          cnt_used_x[i + x] = cnt_used_x[i] + 1;
        }
      }
    }
  }
  if (bs[k]) {
    cout << k << " ";
  } else {
    cout << k + 1 << " ";
  }
}
void solve() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    a[i]--;
  }
  vector<int> lens;
  for (int i = 0; i < n; ++i) {
    if (!used[i]) {
      lens.push_back(1);
      int cur = a[i];
      used[i] = true;
      while (cur != i) {
        used[cur] = true;
        lens.back()++;
        cur = a[cur];
      }
    }
  }
  int cnt_even = 0, cnt_odd = 0;
  for (auto l : lens) {
    if (l % 2) {
      cnt_odd++;
    } else {
      cnt_even++;
    }
  }
  solve1(lens, n, k);
  int k1 = k;
  int ans2 = min(k1, (n - cnt_odd) / 2) * 2;
  k1 -= min(k1, (n - cnt_odd) / 2);
  ans2 += min(k1, cnt_odd);
  cout << ans2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
}

```