#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
const long long INF = 1e18 + 1LL;
const double Pi = acos(-1.0);
const int N = 1e6 + 10, maxn = 1e3 + 20, mod = 1e9 + 7, inf = 2e9;
int a[N], b[N], n, k, vis[N], H[N], cnt;
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]) continue;
    int j = a[i], cnts = 0;
    while (!vis[j]) {
      cnts++;
      vis[j] = 1;
      j = a[j];
    }
    b[++cnt] = cnts;
  }
  sort(b + 1, b + cnt + 1);
  int ans2 = 0, kk = k;
  for (int i = 1; i <= cnt; ++i) {
    if (kk >= b[i] / 2)
      kk -= b[i] / 2, ans2 += b[i] / 2 * 2;
    else {
      ans2 += 2 * kk;
      kk = 0;
      break;
    }
  }
  ans2 += kk;
  ans2 = min(n, ans2);
  for (int i = 1; i <= cnt; ++i) H[b[i]]++;
  bitset<N> dp;
  dp.set(0);
  for (int i = 1; i <= n; ++i) {
    if (H[i] == 0) continue;
    int t = H[i], l = 1;
    while (t) {
      int w = min(t, l);
      t -= w;
      dp |= dp << (w * i);
      l <<= 1;
    }
    if (dp[k] == 1) {
      printf("%d %d\n", k, ans2);
      return 0;
    }
  }
  printf("%d %d\n", 1 + k, ans2);
  return 0;
}

### Prompt

Please create a solution in CPP 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;
#pragma comment(linker, "/STACK:102400000,102400000")
const long long INF = 1e18 + 1LL;
const double Pi = acos(-1.0);
const int N = 1e6 + 10, maxn = 1e3 + 20, mod = 1e9 + 7, inf = 2e9;
int a[N], b[N], n, k, vis[N], H[N], cnt;
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]) continue;
    int j = a[i], cnts = 0;
    while (!vis[j]) {
      cnts++;
      vis[j] = 1;
      j = a[j];
    }
    b[++cnt] = cnts;
  }
  sort(b + 1, b + cnt + 1);
  int ans2 = 0, kk = k;
  for (int i = 1; i <= cnt; ++i) {
    if (kk >= b[i] / 2)
      kk -= b[i] / 2, ans2 += b[i] / 2 * 2;
    else {
      ans2 += 2 * kk;
      kk = 0;
      break;
    }
  }
  ans2 += kk;
  ans2 = min(n, ans2);
  for (int i = 1; i <= cnt; ++i) H[b[i]]++;
  bitset<N> dp;
  dp.set(0);
  for (int i = 1; i <= n; ++i) {
    if (H[i] == 0) continue;
    int t = H[i], l = 1;
    while (t) {
      int w = min(t, l);
      t -= w;
      dp |= dp << (w * i);
      l <<= 1;
    }
    if (dp[k] == 1) {
      printf("%d %d\n", k, ans2);
      return 0;
    }
  }
  printf("%d %d\n", 1 + k, ans2);
  return 0;
}

```