#include <bits/stdc++.h>
using namespace std;
const int N = 1000006;
bool vis[N];
int n, k, a[N];
vector<int> v;
int dfs(int x) {
  vis[x] = 1;
  if (vis[a[x]]) return 1;
  return 1 + dfs(a[x]);
}
bitset<1000006> cur;
map<int, int> mp;
int getmin() {
  cur.set(0);
  for (int x : v) ++mp[x];
  for (auto& t : mp) {
    for (int i = 1; i <= t.second; i <<= 1) {
      t.second -= i;
      cur |= cur << (t.first * i);
      if (cur[k]) return k;
    }
    if (t.second) {
      cur |= cur << (t.first * t.second);
      if (cur[k]) return k;
    }
  }
  return k + 1;
}
int getmax() {
  int tmp, ans = 0;
  for (int& x : v) {
    tmp = min(x >> 1, k);
    ans += tmp << 1;
    k -= tmp;
    x -= tmp << 1;
    if (!k) break;
  }
  if (k) {
    for (int& x : v)
      if (x) {
        ++ans;
        --k;
        if (!k) break;
      }
  }
  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]) v.push_back(dfs(i));
  sort(v.begin(), v.end());
  int m = getmin();
  int M = getmax();
  printf("%d %d\n", m, M);
  return 0;
}

### 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 = 1000006;
bool vis[N];
int n, k, a[N];
vector<int> v;
int dfs(int x) {
  vis[x] = 1;
  if (vis[a[x]]) return 1;
  return 1 + dfs(a[x]);
}
bitset<1000006> cur;
map<int, int> mp;
int getmin() {
  cur.set(0);
  for (int x : v) ++mp[x];
  for (auto& t : mp) {
    for (int i = 1; i <= t.second; i <<= 1) {
      t.second -= i;
      cur |= cur << (t.first * i);
      if (cur[k]) return k;
    }
    if (t.second) {
      cur |= cur << (t.first * t.second);
      if (cur[k]) return k;
    }
  }
  return k + 1;
}
int getmax() {
  int tmp, ans = 0;
  for (int& x : v) {
    tmp = min(x >> 1, k);
    ans += tmp << 1;
    k -= tmp;
    x -= tmp << 1;
    if (!k) break;
  }
  if (k) {
    for (int& x : v)
      if (x) {
        ++ans;
        --k;
        if (!k) break;
      }
  }
  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]) v.push_back(dfs(i));
  sort(v.begin(), v.end());
  int m = getmin();
  int M = getmax();
  printf("%d %d\n", m, M);
  return 0;
}

```