#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 10;
int g[MAXN], t[MAXN];
bool mark[MAXN];
vector<int> c, b;
bitset<MAXN> bs;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> g[i];
  for (int i = 1; i <= n; i++) {
    if (mark[i]) continue;
    int x = i, cnt = 0;
    while (1) {
      if (mark[x]) break;
      mark[x] = 1;
      cnt++;
      x = g[x];
    }
    c.push_back(cnt);
    t[cnt]++;
  }
  sort(c.begin(), c.end());
  reverse(c.begin(), c.end());
  int ans = 0, m = k, o = 0;
  for (auto u : c) {
    if (m >= u / 2) {
      m -= u / 2;
      ans += (u / 2) * 2;
      if (u % 2 == 1) o++;
    } else {
      ans += 2 * m;
      m = 0;
      break;
    }
  }
  ans += min(o, m);
  for (int i = 1; i <= n; i++) {
    if (t[i] == 0) continue;
    int s = 0;
    while ((1 << s) < t[i]) s++;
    if ((1 << s) != t[i]) s--;
    b.push_back(i);
    for (int j = 0; j < s; j++) b.push_back(i * (1 << j));
    b.push_back((t[i] - (1 << s)) * i);
  }
  bs[0] = 1;
  for (auto a : b) bs |= bs << a;
  if (bs[k])
    cout << k;
  else
    cout << k + 1;
  cout << " " << ans;
}

### Prompt

Please provide a CPP coded solution 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 int MAXN = 1e6 + 10;
int g[MAXN], t[MAXN];
bool mark[MAXN];
vector<int> c, b;
bitset<MAXN> bs;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> g[i];
  for (int i = 1; i <= n; i++) {
    if (mark[i]) continue;
    int x = i, cnt = 0;
    while (1) {
      if (mark[x]) break;
      mark[x] = 1;
      cnt++;
      x = g[x];
    }
    c.push_back(cnt);
    t[cnt]++;
  }
  sort(c.begin(), c.end());
  reverse(c.begin(), c.end());
  int ans = 0, m = k, o = 0;
  for (auto u : c) {
    if (m >= u / 2) {
      m -= u / 2;
      ans += (u / 2) * 2;
      if (u % 2 == 1) o++;
    } else {
      ans += 2 * m;
      m = 0;
      break;
    }
  }
  ans += min(o, m);
  for (int i = 1; i <= n; i++) {
    if (t[i] == 0) continue;
    int s = 0;
    while ((1 << s) < t[i]) s++;
    if ((1 << s) != t[i]) s--;
    b.push_back(i);
    for (int j = 0; j < s; j++) b.push_back(i * (1 << j));
    b.push_back((t[i] - (1 << s)) * i);
  }
  bs[0] = 1;
  for (auto a : b) bs |= bs << a;
  if (bs[k])
    cout << k;
  else
    cout << k + 1;
  cout << " " << ans;
}

```