#include <bits/stdc++.h>
using namespace std;
const int INF = (int)2e9;
const long long INFL = (long long)9e18;
const int MAXINT = ((~0) ^ (1 << 31));
const long long MAXLL = ((~0) ^ ((long long)1 << 63));
template <class T>
inline T pow2(T a) {
  return a * a;
}
template <class T>
inline bool mineq(T& a, T b) {
  return (a > b) ? (a = b, true) : false;
}
template <class T>
inline bool maxeq(T& a, T b) {
  return (a < b) ? (a = b, true) : false;
}
const int maxn = 1e6 + 100;
bitset<maxn> dp;
int n, p[maxn], k, cnt[maxn], a, b;
bool vis[maxn];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < (int)(n); i++) {
    cin >> p[i];
    p[i]--;
  }
  for (int i = 0; i < (int)(n); i++) {
    if (vis[i]) continue;
    int tmp = i, len = 0;
    while (!vis[tmp]) {
      vis[tmp] = true;
      len++;
      tmp = p[tmp];
    }
    cnt[len]++;
    a += len % 2;
    b += len / 2;
  }
  dp[0] = true;
  for (int i = 1; i < (int)(n + 1); i++) {
    int j = 1;
    while (cnt[i]) {
      j = min(j, cnt[i]);
      cnt[i] -= j;
      dp |= dp << (i * j);
      j *= 2;
    }
  }
  int mn = k + !dp[k];
  int mx = min(2 * k, min(n, k + b));
  cout << mn << " " << mx << endl;
}

### Prompt

Please formulate 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 INF = (int)2e9;
const long long INFL = (long long)9e18;
const int MAXINT = ((~0) ^ (1 << 31));
const long long MAXLL = ((~0) ^ ((long long)1 << 63));
template <class T>
inline T pow2(T a) {
  return a * a;
}
template <class T>
inline bool mineq(T& a, T b) {
  return (a > b) ? (a = b, true) : false;
}
template <class T>
inline bool maxeq(T& a, T b) {
  return (a < b) ? (a = b, true) : false;
}
const int maxn = 1e6 + 100;
bitset<maxn> dp;
int n, p[maxn], k, cnt[maxn], a, b;
bool vis[maxn];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < (int)(n); i++) {
    cin >> p[i];
    p[i]--;
  }
  for (int i = 0; i < (int)(n); i++) {
    if (vis[i]) continue;
    int tmp = i, len = 0;
    while (!vis[tmp]) {
      vis[tmp] = true;
      len++;
      tmp = p[tmp];
    }
    cnt[len]++;
    a += len % 2;
    b += len / 2;
  }
  dp[0] = true;
  for (int i = 1; i < (int)(n + 1); i++) {
    int j = 1;
    while (cnt[i]) {
      j = min(j, cnt[i]);
      cnt[i] -= j;
      dp |= dp << (i * j);
      j *= 2;
    }
  }
  int mn = k + !dp[k];
  int mx = min(2 * k, min(n, k + b));
  cout << mn << " " << mx << endl;
}

```