#include <bits/stdc++.h>
using namespace std;
int dat[1000005];
int parent[1000005];
int rnk[1000005];
int cnt[1000005];
int cnt2[1000005];
bitset<1000001> bt;
int getparent(int t) {
  return t == parent[t] ? t : (parent[t] = getparent(parent[t]));
}
void merge(int a, int b) {
  a = getparent(a);
  b = getparent(b);
  if (rnk[a] < rnk[b]) swap(a, b);
  if (rnk[a] == rnk[b]) rnk[a]++;
  parent[b] = a;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    parent[i] = i;
  }
  for (int i = 1; i <= n; i++) {
    int t;
    scanf("%d", &t);
    merge(i, t);
  }
  for (int i = 1; i <= n; i++) {
    cnt[getparent(i)]++;
  }
  vector<int> cnts;
  cnts.reserve(n / 2);
  for (int i = 1; i <= n; i++) {
    if (cnt[i]) {
      cnts.push_back(cnt[i]);
      cnt2[cnt[i]]++;
    }
  }
  vector<int> vt;
  for (int i = 1; i <= n; i++) {
    if (cnt2[i]) {
      for (int j = 0; (1 << j) <= cnt2[i]; j++) {
        cnt2[i] -= 1 << j;
        vt.push_back((1 << j) * i);
      }
      if (cnt2[i]) vt.push_back(cnt2[i] * i);
    }
  }
  bt[0] = 1;
  for (int t : vt) bt |= bt << t;
  if (bt[k])
    printf("%d ", k);
  else
    printf("%d ", k + 1);
  int ans2 = 0;
  int remain = k;
  for (int t : cnts) {
    if (remain == 0) break;
    int sub = min(remain, t / 2);
    remain -= sub;
    ans2 += sub * 2;
  }
  for (int t : cnts) {
    if (remain == 0) break;
    if (!(t & 1)) continue;
    remain--;
    ans2++;
  }
  printf("%d", ans2);
}

### 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 dat[1000005];
int parent[1000005];
int rnk[1000005];
int cnt[1000005];
int cnt2[1000005];
bitset<1000001> bt;
int getparent(int t) {
  return t == parent[t] ? t : (parent[t] = getparent(parent[t]));
}
void merge(int a, int b) {
  a = getparent(a);
  b = getparent(b);
  if (rnk[a] < rnk[b]) swap(a, b);
  if (rnk[a] == rnk[b]) rnk[a]++;
  parent[b] = a;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    parent[i] = i;
  }
  for (int i = 1; i <= n; i++) {
    int t;
    scanf("%d", &t);
    merge(i, t);
  }
  for (int i = 1; i <= n; i++) {
    cnt[getparent(i)]++;
  }
  vector<int> cnts;
  cnts.reserve(n / 2);
  for (int i = 1; i <= n; i++) {
    if (cnt[i]) {
      cnts.push_back(cnt[i]);
      cnt2[cnt[i]]++;
    }
  }
  vector<int> vt;
  for (int i = 1; i <= n; i++) {
    if (cnt2[i]) {
      for (int j = 0; (1 << j) <= cnt2[i]; j++) {
        cnt2[i] -= 1 << j;
        vt.push_back((1 << j) * i);
      }
      if (cnt2[i]) vt.push_back(cnt2[i] * i);
    }
  }
  bt[0] = 1;
  for (int t : vt) bt |= bt << t;
  if (bt[k])
    printf("%d ", k);
  else
    printf("%d ", k + 1);
  int ans2 = 0;
  int remain = k;
  for (int t : cnts) {
    if (remain == 0) break;
    int sub = min(remain, t / 2);
    remain -= sub;
    ans2 += sub * 2;
  }
  for (int t : cnts) {
    if (remain == 0) break;
    if (!(t & 1)) continue;
    remain--;
    ans2++;
  }
  printf("%d", ans2);
}

```