#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
const int maxsq = 100;
int cnt[maxn];
int p[maxn], n, k;
bool mark[maxn];
bitset<maxn / 2> bs;
bool tmp[maxn];
int getmin(void) {
  int rem = min(k, n - k);
  bs[0] = 1;
  for (int i = 2; i <= n; ++i)
    if (cnt[i]) {
      if (i <= maxsq) {
        for (int r = 0; r < i; ++r) {
          int su = 0, ptr = 0;
          for (int j = r; j <= k; j += i) {
            if (bs[j]) su++;
            tmp[ptr++] = bs[j];
            bs[j] = bs[j] or su;
            if (j - cnt[i] * i >= 0 and tmp[(j - r) / i - cnt[i]]) su--;
          }
        }
      } else
        for (int j = 0; j < cnt[i]; ++j) bs |= bs << i;
      if (bs[rem]) return k;
    }
  return k + 1;
}
int getmax(void) {
  int ret = 0, rem = k;
  for (int i = 2; i <= n; i += 2) {
    for (int j = 0; j < cnt[i] and rem; j++) {
      if (i / 2 <= rem) {
        ret += i;
        rem -= i / 2;
      } else {
        ret += 2 * rem;
        rem = 0;
      }
    }
  }
  if (!rem) return ret;
  int c = 0;
  for (int i = 1; i <= n; i += 2) {
    for (int j = 0; j < cnt[i] and rem; ++j) {
      if (i / 2 <= rem) {
        ret += i - 1;
        c++;
        rem -= i / 2;
      } else {
        ret += 2 * rem;
        rem = 0;
      }
    }
  }
  return ret + min(c, rem);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; ++i) scanf("%d", p + i), --p[i];
  for (int i = 0; i < n; ++i)
    if (!mark[i]) {
      int u = i;
      int len = 0;
      while (!mark[u]) {
        mark[u] = true;
        len++;
        u = p[u];
      }
      ++cnt[len];
    }
  cout << getmin() << ' ';
  cout << getmax() << '\n';
  return 0;
}

### Prompt

Create a solution in cpp for 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 = 1e6 + 10;
const int maxsq = 100;
int cnt[maxn];
int p[maxn], n, k;
bool mark[maxn];
bitset<maxn / 2> bs;
bool tmp[maxn];
int getmin(void) {
  int rem = min(k, n - k);
  bs[0] = 1;
  for (int i = 2; i <= n; ++i)
    if (cnt[i]) {
      if (i <= maxsq) {
        for (int r = 0; r < i; ++r) {
          int su = 0, ptr = 0;
          for (int j = r; j <= k; j += i) {
            if (bs[j]) su++;
            tmp[ptr++] = bs[j];
            bs[j] = bs[j] or su;
            if (j - cnt[i] * i >= 0 and tmp[(j - r) / i - cnt[i]]) su--;
          }
        }
      } else
        for (int j = 0; j < cnt[i]; ++j) bs |= bs << i;
      if (bs[rem]) return k;
    }
  return k + 1;
}
int getmax(void) {
  int ret = 0, rem = k;
  for (int i = 2; i <= n; i += 2) {
    for (int j = 0; j < cnt[i] and rem; j++) {
      if (i / 2 <= rem) {
        ret += i;
        rem -= i / 2;
      } else {
        ret += 2 * rem;
        rem = 0;
      }
    }
  }
  if (!rem) return ret;
  int c = 0;
  for (int i = 1; i <= n; i += 2) {
    for (int j = 0; j < cnt[i] and rem; ++j) {
      if (i / 2 <= rem) {
        ret += i - 1;
        c++;
        rem -= i / 2;
      } else {
        ret += 2 * rem;
        rem = 0;
      }
    }
  }
  return ret + min(c, rem);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; ++i) scanf("%d", p + i), --p[i];
  for (int i = 0; i < n; ++i)
    if (!mark[i]) {
      int u = i;
      int len = 0;
      while (!mark[u]) {
        mark[u] = true;
        len++;
        u = p[u];
      }
      ++cnt[len];
    }
  cout << getmin() << ' ';
  cout << getmax() << '\n';
  return 0;
}

```