#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000010;
int p[maxn];
bool vis[maxn];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
  }
  vector<int> cir;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      int t = i;
      int len = 0;
      while (!vis[t]) {
        vis[t] = 1;
        t = p[t];
        len++;
      }
      cir.push_back(len);
    }
  }
  sort(cir.begin(), cir.end());
  int MAX = 0;
  int kk = k;
  for (int i = 0; i < cir.size(); i++) {
    int tmp = cir[i];
    while (tmp >= 2 && kk > 0) {
      MAX += 2;
      kk--;
      tmp -= 2;
    }
  }
  MAX += kk;
  MAX = min(MAX, n);
  vector<int> vec;
  cir.push_back(-1);
  int s = 0;
  for (int i = 1; i < cir.size(); i++) {
    while (cir[i] == cir[i - 1]) {
      i++;
    }
    int t = i - s;
    s = i;
    int j = 0;
    while (t) {
      int tmp = min(1 << j, t);
      t -= tmp;
      j++;
      vec.push_back(cir[i - 1] * tmp);
    }
    if (cir[i] == -1) {
      break;
    }
  }
  bitset<maxn> bs[2];
  int cur = 0, pre = 1;
  bs[cur][0] = 1;
  for (int i = 0; i < vec.size(); i++) {
    swap(cur, pre);
    bs[cur] = bs[pre] | (bs[pre] << vec[i]);
  }
  int MIN = k + 1 - bs[cur][k];
  cout << MIN << " " << MAX << endl;
  return 0;
}

### Prompt

Your challenge is to write 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 maxn = 1000010;
int p[maxn];
bool vis[maxn];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
  }
  vector<int> cir;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      int t = i;
      int len = 0;
      while (!vis[t]) {
        vis[t] = 1;
        t = p[t];
        len++;
      }
      cir.push_back(len);
    }
  }
  sort(cir.begin(), cir.end());
  int MAX = 0;
  int kk = k;
  for (int i = 0; i < cir.size(); i++) {
    int tmp = cir[i];
    while (tmp >= 2 && kk > 0) {
      MAX += 2;
      kk--;
      tmp -= 2;
    }
  }
  MAX += kk;
  MAX = min(MAX, n);
  vector<int> vec;
  cir.push_back(-1);
  int s = 0;
  for (int i = 1; i < cir.size(); i++) {
    while (cir[i] == cir[i - 1]) {
      i++;
    }
    int t = i - s;
    s = i;
    int j = 0;
    while (t) {
      int tmp = min(1 << j, t);
      t -= tmp;
      j++;
      vec.push_back(cir[i - 1] * tmp);
    }
    if (cir[i] == -1) {
      break;
    }
  }
  bitset<maxn> bs[2];
  int cur = 0, pre = 1;
  bs[cur][0] = 1;
  for (int i = 0; i < vec.size(); i++) {
    swap(cur, pre);
    bs[cur] = bs[pre] | (bs[pre] << vec[i]);
  }
  int MIN = k + 1 - bs[cur][k];
  cout << MIN << " " << MAX << endl;
  return 0;
}

```