#include <bits/stdc++.h>
using namespace std;
const int NK = 1e6 + 10;
int p[NK], qtd[NK];
bool seen[NK];
int dp[NK];
int get_minimum(int n, int k) {
  int lim = min(k, n - k);
  vector<pair<int, int>> cycles;
  for (int i = 1; i <= n; i++) {
    if (qtd[i]) cycles.push_back({i, qtd[i]});
  }
  dp[0] = 0;
  for (int i = 1; i <= lim; i++) {
    dp[i] = -1;
  }
  for (auto c : cycles) {
    int val = c.first, qtd = c.second;
    for (int i = 0; i <= lim; i++) {
      if (dp[i] != -1) {
        dp[i] = 0;
      } else if (i >= val && dp[i - val] != -1 && dp[i - val] < qtd) {
        dp[i] = dp[i - val] + 1;
      } else {
        dp[i] = -1;
      }
    }
  }
  return dp[lim] != -1 ? k : k + 1;
}
int get_maximum(int n, int k) {
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    cnt += i / 2 * qtd[i];
  }
  return min(min(cnt, k) * 2 + k - min(cnt, k), n);
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
  }
  for (int i = 1; i <= n; i++)
    if (!seen[i]) {
      int u = i, cnt = 0;
      while (!seen[u]) {
        seen[u] = true;
        u = p[u];
        cnt++;
      }
      qtd[cnt]++;
    }
  printf("%d %d\n", get_minimum(n, k), get_maximum(n, k));
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 NK = 1e6 + 10;
int p[NK], qtd[NK];
bool seen[NK];
int dp[NK];
int get_minimum(int n, int k) {
  int lim = min(k, n - k);
  vector<pair<int, int>> cycles;
  for (int i = 1; i <= n; i++) {
    if (qtd[i]) cycles.push_back({i, qtd[i]});
  }
  dp[0] = 0;
  for (int i = 1; i <= lim; i++) {
    dp[i] = -1;
  }
  for (auto c : cycles) {
    int val = c.first, qtd = c.second;
    for (int i = 0; i <= lim; i++) {
      if (dp[i] != -1) {
        dp[i] = 0;
      } else if (i >= val && dp[i - val] != -1 && dp[i - val] < qtd) {
        dp[i] = dp[i - val] + 1;
      } else {
        dp[i] = -1;
      }
    }
  }
  return dp[lim] != -1 ? k : k + 1;
}
int get_maximum(int n, int k) {
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    cnt += i / 2 * qtd[i];
  }
  return min(min(cnt, k) * 2 + k - min(cnt, k), n);
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
  }
  for (int i = 1; i <= n; i++)
    if (!seen[i]) {
      int u = i, cnt = 0;
      while (!seen[u]) {
        seen[u] = true;
        u = p[u];
        cnt++;
      }
      qtd[cnt]++;
    }
  printf("%d %d\n", get_minimum(n, k), get_maximum(n, k));
  return 0;
}

```