#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 10;
int N, K;
int A[MAXN];
bool vis[MAXN];
vector<int> vlen;
int cnt[MAXN];
int getmin() {
  bitset<MAXN> b;
  b[0] = true;
  for (int i = 1; i <= N; i++) {
    for (int c = cnt[i], step = 1; c; step *= 2) {
      step = min(step, c);
      b |= (b << (i * step));
      c -= step;
    }
  }
  return K + !b[K];
}
int getmax() {
  int cnt1 = 0, cnt2 = 0;
  for (int x : vlen) {
    cnt1 += x % 2;
    cnt2 += x / 2;
  }
  int ck = K;
  int ans = 0;
  ans += 2 * min(ck, cnt2);
  ck -= min(ck, cnt2);
  ans += min(ck, cnt1);
  return ans;
}
int main() {
  scanf("%d %d", &N, &K);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &A[i]);
  }
  for (int i = 1; i <= N; i++) {
    if (vis[i]) {
      continue;
    }
    int clen = 0;
    for (int j = i; !vis[j]; j = A[j]) {
      clen++;
      vis[j] = true;
    }
    cnt[clen]++;
    vlen.push_back(clen);
  }
  printf("%d %d\n", getmin(), getmax());
}

### Prompt

Your task is to create 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;
const int MAXN = 1e6 + 10;
int N, K;
int A[MAXN];
bool vis[MAXN];
vector<int> vlen;
int cnt[MAXN];
int getmin() {
  bitset<MAXN> b;
  b[0] = true;
  for (int i = 1; i <= N; i++) {
    for (int c = cnt[i], step = 1; c; step *= 2) {
      step = min(step, c);
      b |= (b << (i * step));
      c -= step;
    }
  }
  return K + !b[K];
}
int getmax() {
  int cnt1 = 0, cnt2 = 0;
  for (int x : vlen) {
    cnt1 += x % 2;
    cnt2 += x / 2;
  }
  int ck = K;
  int ans = 0;
  ans += 2 * min(ck, cnt2);
  ck -= min(ck, cnt2);
  ans += min(ck, cnt1);
  return ans;
}
int main() {
  scanf("%d %d", &N, &K);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &A[i]);
  }
  for (int i = 1; i <= N; i++) {
    if (vis[i]) {
      continue;
    }
    int clen = 0;
    for (int j = i; !vis[j]; j = A[j]) {
      clen++;
      vis[j] = true;
    }
    cnt[clen]++;
    vlen.push_back(clen);
  }
  printf("%d %d\n", getmin(), getmax());
}

```