#include <bits/stdc++.h>
using namespace std;
int n, k, ile2, ile1, Max;
vector<int> r;
int p[1000007];
bitset<1000007> vis;
bitset<1000007> dp;
queue<pair<int, int> > Q;
int ile[1000007];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> p[i];
  }
  for (int i = 1; i <= n; ++i) {
    if (!vis[i]) {
      int curr = 1, v = i;
      vis[v] = 1;
      while (!vis[p[v]]) {
        ++curr;
        v = p[v];
        vis[v] = 1;
      }
      r.push_back(curr);
    }
  }
  for (auto it : r) {
    ile2 += it / 2;
    ile1 += it % 2;
  }
  int curr = k;
  while (curr > 0 && ile2 > 0) {
    --ile2;
    --curr;
    Max += 2;
  }
  while (curr > 0 && ile1 > 0) {
    --ile1;
    --curr;
    Max += 1;
  }
  for (auto it : r) ++ile[it];
  for (int i = 1; i <= n; ++i) {
    if (ile[i] > 2) {
      ile[2 * i] += (ile[i] - 1) / 2;
      ile[i] -= (ile[i] - 1) / 2 * 2;
    }
  }
  dp[0] = 1;
  for (int i = 1; i <= n; ++i) {
    while (ile[i]--) dp = dp | (dp << i);
  }
  if (dp[k])
    cout << k << " " << Max;
  else
    cout << k + 1 << " " << Max;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 n, k, ile2, ile1, Max;
vector<int> r;
int p[1000007];
bitset<1000007> vis;
bitset<1000007> dp;
queue<pair<int, int> > Q;
int ile[1000007];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> p[i];
  }
  for (int i = 1; i <= n; ++i) {
    if (!vis[i]) {
      int curr = 1, v = i;
      vis[v] = 1;
      while (!vis[p[v]]) {
        ++curr;
        v = p[v];
        vis[v] = 1;
      }
      r.push_back(curr);
    }
  }
  for (auto it : r) {
    ile2 += it / 2;
    ile1 += it % 2;
  }
  int curr = k;
  while (curr > 0 && ile2 > 0) {
    --ile2;
    --curr;
    Max += 2;
  }
  while (curr > 0 && ile1 > 0) {
    --ile1;
    --curr;
    Max += 1;
  }
  for (auto it : r) ++ile[it];
  for (int i = 1; i <= n; ++i) {
    if (ile[i] > 2) {
      ile[2 * i] += (ile[i] - 1) / 2;
      ile[i] -= (ile[i] - 1) / 2 * 2;
    }
  }
  dp[0] = 1;
  for (int i = 1; i <= n; ++i) {
    while (ile[i]--) dp = dp | (dp << i);
  }
  if (dp[k])
    cout << k << " " << Max;
  else
    cout << k + 1 << " " << Max;
}

```