#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int n, p[N], k, cnt[N], t2[N];
bitset<N> dp;
vector<int> A;
bool mark[N];
void PREP() {
  int x, t;
  for (int i = 1; i <= n; i++) {
    if (mark[i]) continue;
    x = i, t = 0;
    while (!mark[x]) mark[x] = 1, t++, x = p[x];
    cnt[t]++;
  }
  x = 1;
  while (x <= n) t2[x] = x, x *= 2;
  for (int i = 1; i <= n; i++) {
    if (t2[i])
      x = i;
    else
      t2[i] = x;
  }
  for (int i = 1; i <= n; i++) {
    x = 1, t = t2[cnt[i]];
    if (cnt[i]) {
      A.push_back(x * i);
      while (x < t) A.push_back(x * i), x *= 2;
      if (cnt[i] - t > 0) A.push_back(i * (cnt[i] - t2[cnt[i]]));
    }
  }
}
void MIN(int k) {
  dp[0] = 1;
  for (auto x : A) dp = (dp << x) | dp;
  if (!dp[k]) k++;
  printf("%d ", k);
  return;
}
void MAX(int k) {
  int res = 0;
  for (int i = n; i; --i) {
    int q = cnt[i];
    while (k && q--) {
      if (k - i / 2 < 0)
        res += 2 * k, k = 0;
      else
        res += i / 2 * 2, k -= i / 2;
    }
  }
  for (int i = n; i; --i) {
    int q = cnt[i];
    while (i & 1 && k && q--) k--, res++;
  }
  printf("%d ", res);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  PREP();
  MIN(k);
  MAX(k);
}

### 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;
const int N = 1e6 + 10;
int n, p[N], k, cnt[N], t2[N];
bitset<N> dp;
vector<int> A;
bool mark[N];
void PREP() {
  int x, t;
  for (int i = 1; i <= n; i++) {
    if (mark[i]) continue;
    x = i, t = 0;
    while (!mark[x]) mark[x] = 1, t++, x = p[x];
    cnt[t]++;
  }
  x = 1;
  while (x <= n) t2[x] = x, x *= 2;
  for (int i = 1; i <= n; i++) {
    if (t2[i])
      x = i;
    else
      t2[i] = x;
  }
  for (int i = 1; i <= n; i++) {
    x = 1, t = t2[cnt[i]];
    if (cnt[i]) {
      A.push_back(x * i);
      while (x < t) A.push_back(x * i), x *= 2;
      if (cnt[i] - t > 0) A.push_back(i * (cnt[i] - t2[cnt[i]]));
    }
  }
}
void MIN(int k) {
  dp[0] = 1;
  for (auto x : A) dp = (dp << x) | dp;
  if (!dp[k]) k++;
  printf("%d ", k);
  return;
}
void MAX(int k) {
  int res = 0;
  for (int i = n; i; --i) {
    int q = cnt[i];
    while (k && q--) {
      if (k - i / 2 < 0)
        res += 2 * k, k = 0;
      else
        res += i / 2 * 2, k -= i / 2;
    }
  }
  for (int i = n; i; --i) {
    int q = cnt[i];
    while (i & 1 && k && q--) k--, res++;
  }
  printf("%d ", res);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  PREP();
  MIN(k);
  MAX(k);
}

```