#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;
int p[N], mark[N], cnt[N];
vector<int> vec;
bitset<N> dp;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> p[i];
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    if (mark[i]) continue;
    int v = i, len = 0;
    while (!mark[v]) mark[v] = 1, len++, v = p[v];
    cnt[len]++;
    sum += len - len % 2;
  }
  int mx = min(n, k + min(k, sum / 2));
  for (int i = 1; i <= n; i++) {
    int pw = 1;
    while (pw <= cnt[i]) vec.push_back(i * pw), cnt[i] -= pw, pw *= 2;
    if (cnt[i]) vec.push_back(i * cnt[i]);
  }
  int cp = k;
  k = min(k, n - k);
  dp[0] = 1;
  for (int x : vec) {
    dp |= dp << x;
    if (dp[k]) break;
  }
  int mn = cp + 1 - dp[k];
  cout << mn << ' ' << mx << '\n';
  return 0;
}

### Prompt

Develop a solution in Cpp 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 N = 1000005;
int p[N], mark[N], cnt[N];
vector<int> vec;
bitset<N> dp;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> p[i];
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    if (mark[i]) continue;
    int v = i, len = 0;
    while (!mark[v]) mark[v] = 1, len++, v = p[v];
    cnt[len]++;
    sum += len - len % 2;
  }
  int mx = min(n, k + min(k, sum / 2));
  for (int i = 1; i <= n; i++) {
    int pw = 1;
    while (pw <= cnt[i]) vec.push_back(i * pw), cnt[i] -= pw, pw *= 2;
    if (cnt[i]) vec.push_back(i * cnt[i]);
  }
  int cp = k;
  k = min(k, n - k);
  dp[0] = 1;
  for (int x : vec) {
    dp |= dp << x;
    if (dp[k]) break;
  }
  int mn = cp + 1 - dp[k];
  cout << mn << ' ' << mx << '\n';
  return 0;
}

```