#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
int a[1000006];
bitset<1000006> v, can;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  vector<int> szs;
  for (int i = 1; i <= n; ++i) {
    if (v[i]) continue;
    szs.push_back(1);
    v[i] = 1;
    int j = a[i];
    while (j != i) {
      v[j] = 1;
      ++szs.back();
      j = a[j];
    }
  }
  [&]() {
    can[0] = 1;
    vector<int> big, small(1005, 0);
    for (int i : szs) {
      if (i <= 1000)
        ++small[i];
      else
        big.push_back(i);
    }
    for (int i : big) can |= can << i;
    for (int i = 1, j; i <= 1000; ++i) {
      for (j = 0; (1 << j) <= small[i]; ++j) {
        can |= can << (i * (1 << j));
        small[i] -= (1 << j);
      }
      can |= can << (i * (small[i]));
    }
    if (can[k])
      printf("%d ", k);
    else
      printf("%d ", k + 1);
  }();
  [&, k]() mutable {
    int c[5] = {0};
    for (int i : szs) {
      while (i > 1) {
        i -= 2;
        ++c[2];
      }
      if (i) ++c[1];
    }
    int ans = 0;
    if (c[2] < k) {
      ans += c[2] * 2;
      k -= c[2];
      if (c[1] < k)
        ans += c[1];
      else
        ans += k;
    } else
      ans += k * 2;
    printf("%d", ans);
  }();
}

### Prompt

In CPP, your task is to solve 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>
#pragma GCC optimize("Ofast")
using namespace std;
int a[1000006];
bitset<1000006> v, can;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  vector<int> szs;
  for (int i = 1; i <= n; ++i) {
    if (v[i]) continue;
    szs.push_back(1);
    v[i] = 1;
    int j = a[i];
    while (j != i) {
      v[j] = 1;
      ++szs.back();
      j = a[j];
    }
  }
  [&]() {
    can[0] = 1;
    vector<int> big, small(1005, 0);
    for (int i : szs) {
      if (i <= 1000)
        ++small[i];
      else
        big.push_back(i);
    }
    for (int i : big) can |= can << i;
    for (int i = 1, j; i <= 1000; ++i) {
      for (j = 0; (1 << j) <= small[i]; ++j) {
        can |= can << (i * (1 << j));
        small[i] -= (1 << j);
      }
      can |= can << (i * (small[i]));
    }
    if (can[k])
      printf("%d ", k);
    else
      printf("%d ", k + 1);
  }();
  [&, k]() mutable {
    int c[5] = {0};
    for (int i : szs) {
      while (i > 1) {
        i -= 2;
        ++c[2];
      }
      if (i) ++c[1];
    }
    int ans = 0;
    if (c[2] < k) {
      ans += c[2] * 2;
      k -= c[2];
      if (c[1] < k)
        ans += c[1];
      else
        ans += k;
    } else
      ans += k * 2;
    printf("%d", ans);
  }();
}

```