#include <bits/stdc++.h>
using namespace std;
int P[1000005];
bool used[1000005];
map<int, int> num;
bitset<1000005> can;
int main() {
  int N, K;
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &P[i]);
  }
  vector<int> sizs;
  for (int i = 1; i <= N; ++i) {
    if (used[i]) continue;
    int siz = 1;
    used[i] = true;
    for (int x = P[i]; x != i; x = P[x]) {
      used[x] = true;
      ++siz;
    }
    sizs.push_back(siz);
    ++num[siz];
  }
  int n = sizs.size();
  int mn;
  can[0] = 1;
  for (auto y : num) {
    int x = y.first;
    int nn = y.second;
    int w = 1;
    while (nn) {
      w = min(w, nn);
      can |= can << (x * w);
      nn -= w;
      w *= 2;
    }
  }
  if (!K)
    mn = 0;
  else if (can[K])
    mn = K;
  else
    mn = K + 1;
  int mx, mx2s = 0;
  for (int x : sizs) {
    mx2s += x / 2;
  }
  if (K <= mx2s)
    mx = K * 2;
  else
    mx = min(N, mx2s * 2 + (K - mx2s));
  printf("%d %d\n", mn, mx);
}

### 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;
int P[1000005];
bool used[1000005];
map<int, int> num;
bitset<1000005> can;
int main() {
  int N, K;
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &P[i]);
  }
  vector<int> sizs;
  for (int i = 1; i <= N; ++i) {
    if (used[i]) continue;
    int siz = 1;
    used[i] = true;
    for (int x = P[i]; x != i; x = P[x]) {
      used[x] = true;
      ++siz;
    }
    sizs.push_back(siz);
    ++num[siz];
  }
  int n = sizs.size();
  int mn;
  can[0] = 1;
  for (auto y : num) {
    int x = y.first;
    int nn = y.second;
    int w = 1;
    while (nn) {
      w = min(w, nn);
      can |= can << (x * w);
      nn -= w;
      w *= 2;
    }
  }
  if (!K)
    mn = 0;
  else if (can[K])
    mn = K;
  else
    mn = K + 1;
  int mx, mx2s = 0;
  for (int x : sizs) {
    mx2s += x / 2;
  }
  if (K <= mx2s)
    mx = K * 2;
  else
    mx = min(N, mx2s * 2 + (K - mx2s));
  printf("%d %d\n", mn, mx);
}

```