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.
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:32000000")
using namespace std;
inline bool read(int& val) { return scanf("%d", &val) != -1; }
inline bool read(long long& val) { return scanf("%I64d", &val) != -1; }
inline bool read(double& val) { return scanf("%lf", &val) != -1; }
inline bool read(char* val) { return scanf("%s", val) != -1; }
template <class T1, class T2>
inline bool read(T1& a, T2& b) {
  return read(a) && read(b);
}
template <class T1, class T2, class T3>
inline bool read(T1& a, T2& b, T3& c) {
  return read(a) && read(b) && read(c);
}
template <class T1, class T2, class T3, class T4>
inline bool read(T1& a, T2& b, T3& c, T4& d) {
  return read(a) && read(b) && read(c) && read(d);
}
template <class T1, class T2, class T3, class T4, class T5>
inline bool read(T1& a, T2& b, T3& c, T4& d, T5& e) {
  return read(a) && read(b) && read(c) && read(d) && read(e);
}
const int N = 1111111;
const long long mod = 1000000007;
int A[N];
int cnt[N];
bool vis[N];
bitset<N> dp;
int main() {
  int n, k;
  read(n, k);
  for (int i = 1, _ = (n); i <= _; ++i) read(A[i]);
  int odd = 0;
  for (int i = 1, _ = (n); i <= _; ++i) {
    int len = 0;
    for (int j = i; !vis[j]; j = A[j]) len += (vis[j] = true);
    cnt[len]++;
    odd += len % 2;
  }
  for (int i = 1, _ = (n); i <= _; ++i)
    while (cnt[i] > 2) {
      cnt[i * ((cnt[i] + 1) / 2)]++;
      cnt[i] /= 2;
    }
  dp[0] = 1;
  for (int i = 1, _ = (n); i <= _; ++i)
    for (int j = 0, _ = (cnt[i]); j < _; ++j) dp |= dp << i;
  cout << (dp[k] ? k : k + 1) << ' ';
  if (k * 2 <= n - odd)
    cout << k * 2 << endl;
  else
    cout << min(n, k + (n - odd) / 2) << endl;
  return 0;
}
