#include <bits/stdc++.h>
using namespace std;
int N, K, p[1100100], sz[1100100], cnt[1500500], mn, mx, numfree, numone;
vector<int> o, e, v;
int req[1100100];
bool dp[2200200];
int g(int i) { return p[i] == i ? i : p[i] = g(p[i]); }
int main() {
  scanf("%d %d", &N, &K);
  for (int i = (1); i < (N + 1); i++) p[i] = i;
  for (int i = (1); i < (N + 1); i++) {
    int x;
    scanf("%d", &x);
    p[g(x)] = g(i);
  }
  for (int i = (1); i < (N + 1); i++) sz[g(i)]++;
  for (int i = (1); i < (N + 1); i++)
    if (sz[i]) {
      if (sz[i] & 1)
        o.push_back(sz[i]), numfree += sz[i] / 2, numone++;
      else
        e.push_back(sz[i]), numfree += sz[i] / 2;
      cnt[sz[i]]++;
      v.push_back(sz[i]);
    }
  sort(v.begin(), v.end());
  reverse(v.begin(), v.end());
  dp[0] = 1;
  for (int i = (1); i < (N + 1); i++)
    if (cnt[i]) {
      for (int j = 0; j + i <= N; j++) {
        if (dp[j] && !dp[j + i] && req[j] < cnt[i]) {
          dp[j + i] = 1, req[j + i] = req[j] + 1;
        }
        req[j] = 0;
      }
    }
  mn = dp[K] ? K : K + 1;
  numfree = min(numfree, K);
  K -= numfree;
  mx += numfree * 2;
  mx += min(K, numone);
  printf("%d %d\n", mn, mx);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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, p[1100100], sz[1100100], cnt[1500500], mn, mx, numfree, numone;
vector<int> o, e, v;
int req[1100100];
bool dp[2200200];
int g(int i) { return p[i] == i ? i : p[i] = g(p[i]); }
int main() {
  scanf("%d %d", &N, &K);
  for (int i = (1); i < (N + 1); i++) p[i] = i;
  for (int i = (1); i < (N + 1); i++) {
    int x;
    scanf("%d", &x);
    p[g(x)] = g(i);
  }
  for (int i = (1); i < (N + 1); i++) sz[g(i)]++;
  for (int i = (1); i < (N + 1); i++)
    if (sz[i]) {
      if (sz[i] & 1)
        o.push_back(sz[i]), numfree += sz[i] / 2, numone++;
      else
        e.push_back(sz[i]), numfree += sz[i] / 2;
      cnt[sz[i]]++;
      v.push_back(sz[i]);
    }
  sort(v.begin(), v.end());
  reverse(v.begin(), v.end());
  dp[0] = 1;
  for (int i = (1); i < (N + 1); i++)
    if (cnt[i]) {
      for (int j = 0; j + i <= N; j++) {
        if (dp[j] && !dp[j + i] && req[j] < cnt[i]) {
          dp[j + i] = 1, req[j + i] = req[j] + 1;
        }
        req[j] = 0;
      }
    }
  mn = dp[K] ? K : K + 1;
  numfree = min(numfree, K);
  K -= numfree;
  mx += numfree * 2;
  mx += min(K, numone);
  printf("%d %d\n", mn, mx);
  return 0;
}

```