#include <bits/stdc++.h>
using namespace std;
int Q, N, M, K, L;
bitset<1000001> BITSET;
int p[1000001];
bool chk[1000001];
int cycles[1000001];
int clen = 0;
int main() {
  ios::sync_with_stdio(false);
  cin >> N >> M;
  int oc = 0, ec = 0;
  for (int i = 0; i < N; i++) cin >> p[i], p[i]--;
  for (int i = 0; i < N; i++) {
    if (chk[i]) continue;
    int cur = i;
    int r = 0;
    while (1) {
      chk[cur] = 1;
      cur = p[cur];
      r++;
      if (cur == i) break;
    }
    cycles[r]++;
    ec += r / 2;
    oc += r % 2;
  }
  int maxn = min(M, ec) * 2;
  int maxnr = M - min(M, ec);
  maxn += min(oc, maxnr);
  BITSET.set(0);
  for (int i = 0; i < 1000001; i++) {
    int k = 1;
    while (cycles[i]) {
      if (cycles[i] < k) k = 1;
      BITSET |= BITSET << k * i;
      cycles[i] -= k;
      k *= 2;
    }
  }
  if (BITSET.test(M))
    cout << M;
  else
    cout << M + 1;
  cout << " " << maxn;
  cin >> N;
}

### Prompt

Generate 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 Q, N, M, K, L;
bitset<1000001> BITSET;
int p[1000001];
bool chk[1000001];
int cycles[1000001];
int clen = 0;
int main() {
  ios::sync_with_stdio(false);
  cin >> N >> M;
  int oc = 0, ec = 0;
  for (int i = 0; i < N; i++) cin >> p[i], p[i]--;
  for (int i = 0; i < N; i++) {
    if (chk[i]) continue;
    int cur = i;
    int r = 0;
    while (1) {
      chk[cur] = 1;
      cur = p[cur];
      r++;
      if (cur == i) break;
    }
    cycles[r]++;
    ec += r / 2;
    oc += r % 2;
  }
  int maxn = min(M, ec) * 2;
  int maxnr = M - min(M, ec);
  maxn += min(oc, maxnr);
  BITSET.set(0);
  for (int i = 0; i < 1000001; i++) {
    int k = 1;
    while (cycles[i]) {
      if (cycles[i] < k) k = 1;
      BITSET |= BITSET << k * i;
      cycles[i] -= k;
      k *= 2;
    }
  }
  if (BITSET.test(M))
    cout << M;
  else
    cout << M + 1;
  cout << " " << maxn;
  cin >> N;
}

```