#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int n, k;
bool h[N];
int a[N], cyc[N], p[N];
bitset<N> dp;
void create() {
  memset(h, 0, sizeof(h));
  memset(cyc, 0, sizeof(cyc));
  memset(p, 0, sizeof(p));
  for (int i = 1; i <= n; i++) {
    if (!h[i]) {
      int x = i, cnt = 0;
      while (!h[x]) {
        h[x] = 1;
        cnt++;
        x = a[x];
      }
      cyc[cnt]++;
    }
  }
}
int main() {
  scanf("%d %d", &n, &k);
  int kk = k;
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  int mn = n;
  create();
  for (int i = n; i >= 2; i--) {
    while (k and cyc[i]) {
      k--;
      cyc[i]--;
      mn -= 2;
      p[i - 1]++;
    }
    while (k and i >= 3 and p[i]) {
      k--;
      p[i]--;
      mn -= 2;
      p[i - 2]++;
    }
    while (k and i == 2 and p[i]) {
      k--;
      p[i]--;
      mn -= 1;
    }
  }
  create();
  k = kk;
  int mx = n - k;
  dp[0] = 1;
  for (int i = 1; i <= n; i++) {
    if (!cyc[i]) continue;
    int rem = cyc[i];
    for (int j = 1; j * 2 - 1 <= cyc[i]; j *= 2) {
      dp |= dp << (j * i);
      rem = cyc[i] - (j * 2 - 1);
    }
    if (rem) {
      dp |= dp << (rem * i);
    }
  }
  mx -= !dp[k];
  printf("%d %d\n", n - mx, n - mn);
  return 0;
}

### Prompt

In CPP, your task is to solve 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 N = 1e6 + 5;
int n, k;
bool h[N];
int a[N], cyc[N], p[N];
bitset<N> dp;
void create() {
  memset(h, 0, sizeof(h));
  memset(cyc, 0, sizeof(cyc));
  memset(p, 0, sizeof(p));
  for (int i = 1; i <= n; i++) {
    if (!h[i]) {
      int x = i, cnt = 0;
      while (!h[x]) {
        h[x] = 1;
        cnt++;
        x = a[x];
      }
      cyc[cnt]++;
    }
  }
}
int main() {
  scanf("%d %d", &n, &k);
  int kk = k;
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  int mn = n;
  create();
  for (int i = n; i >= 2; i--) {
    while (k and cyc[i]) {
      k--;
      cyc[i]--;
      mn -= 2;
      p[i - 1]++;
    }
    while (k and i >= 3 and p[i]) {
      k--;
      p[i]--;
      mn -= 2;
      p[i - 2]++;
    }
    while (k and i == 2 and p[i]) {
      k--;
      p[i]--;
      mn -= 1;
    }
  }
  create();
  k = kk;
  int mx = n - k;
  dp[0] = 1;
  for (int i = 1; i <= n; i++) {
    if (!cyc[i]) continue;
    int rem = cyc[i];
    for (int j = 1; j * 2 - 1 <= cyc[i]; j *= 2) {
      dp |= dp << (j * i);
      rem = cyc[i] - (j * 2 - 1);
    }
    if (rem) {
      dp |= dp << (rem * i);
    }
  }
  mx -= !dp[k];
  printf("%d %d\n", n - mx, n - mn);
  return 0;
}

```