#include <bits/stdc++.h>
using namespace std;
int n, k;
int a[1000007];
vector<int> v;
int used[1000007];
bitset<1000001> dp;
int lft[1000007];
int tot[700 + 1];
int solve_min() {
  int sz = v.size();
  dp[0] = true;
  int sm = 0;
  for (int i = 0; i < sz; ++i) {
    if (dp[k] == true) {
      break;
    }
    sm += v[i];
    if (v[i] >= 700) {
      dp |= (dp << v[i]);
    }
  }
  for (int i = 0; i < sz; ++i) {
    if (v[i] < 700) {
      if (tot[v[i]] == 0) {
        continue;
      }
      for (int j = 0; j < v[i]; ++j) {
        if (dp[j] == true) {
          lft[j] = tot[v[i]];
        } else {
          lft[j] = 0;
        }
      }
      for (int j = v[i]; j <= sm; ++j) {
        if (dp[j] == true) {
          lft[j] = tot[v[i]];
        } else {
          if (lft[j - v[i]] > 0) {
            dp[j] = true;
            lft[j] = lft[j - v[i]] - 1;
          } else {
            lft[j] = 0;
          }
        }
      }
      tot[v[i]] = 0;
    }
  }
  if (dp[k] == true) {
    return k;
  }
  return k + 1;
}
int solve_max() {
  int sz = v.size();
  int ret = 0;
  int aux = k;
  for (int i = 0; i < sz; ++i) {
    int x = v[i];
    while (aux > 0 && x >= 2) {
      --aux;
      x -= 2;
      ret += 2;
    }
  }
  return min(n, ret + aux);
}
void input() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
}
void solve() {
  for (int i = 1; i <= n; ++i) {
    if (used[i] == 1) {
      continue;
    }
    int x = i;
    int cnt = 0;
    while (1) {
      used[x] = 1;
      x = a[x];
      ++cnt;
      if (x == i) {
        break;
      }
    }
    v.push_back(cnt);
    if (cnt < 700) {
      ++tot[cnt];
    }
  }
  printf("%d %d\n", solve_min(), solve_max());
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  solve();
  return 0;
}

### 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;
int n, k;
int a[1000007];
vector<int> v;
int used[1000007];
bitset<1000001> dp;
int lft[1000007];
int tot[700 + 1];
int solve_min() {
  int sz = v.size();
  dp[0] = true;
  int sm = 0;
  for (int i = 0; i < sz; ++i) {
    if (dp[k] == true) {
      break;
    }
    sm += v[i];
    if (v[i] >= 700) {
      dp |= (dp << v[i]);
    }
  }
  for (int i = 0; i < sz; ++i) {
    if (v[i] < 700) {
      if (tot[v[i]] == 0) {
        continue;
      }
      for (int j = 0; j < v[i]; ++j) {
        if (dp[j] == true) {
          lft[j] = tot[v[i]];
        } else {
          lft[j] = 0;
        }
      }
      for (int j = v[i]; j <= sm; ++j) {
        if (dp[j] == true) {
          lft[j] = tot[v[i]];
        } else {
          if (lft[j - v[i]] > 0) {
            dp[j] = true;
            lft[j] = lft[j - v[i]] - 1;
          } else {
            lft[j] = 0;
          }
        }
      }
      tot[v[i]] = 0;
    }
  }
  if (dp[k] == true) {
    return k;
  }
  return k + 1;
}
int solve_max() {
  int sz = v.size();
  int ret = 0;
  int aux = k;
  for (int i = 0; i < sz; ++i) {
    int x = v[i];
    while (aux > 0 && x >= 2) {
      --aux;
      x -= 2;
      ret += 2;
    }
  }
  return min(n, ret + aux);
}
void input() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
}
void solve() {
  for (int i = 1; i <= n; ++i) {
    if (used[i] == 1) {
      continue;
    }
    int x = i;
    int cnt = 0;
    while (1) {
      used[x] = 1;
      x = a[x];
      ++cnt;
      if (x == i) {
        break;
      }
    }
    v.push_back(cnt);
    if (cnt < 700) {
      ++tot[cnt];
    }
  }
  printf("%d %d\n", solve_min(), solve_max());
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  solve();
  return 0;
}

```