#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void read(T& num) {
  num = 0;
  bool f = true;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = false;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    num = num * 10 + ch - '0';
    ch = getchar();
  }
  num = f ? num : -num;
}
int out[100];
template <class T>
inline void write(T x, char ch) {
  if (x == 0) {
    putchar('0');
    putchar(ch);
    return;
  }
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  int num = 0;
  while (x) {
    out[num++] = (x % 10);
    x = x / 10;
  }
  for (int i = (num - 1); i >= (0); i--) putchar(out[i] + '0');
  putchar(ch);
}
const double eps = 1e-8;
bitset<1000010> f;
bool vis[1000005];
int c[1000005], cc[1000005], nex[1000005];
int n, k, cnt;
int main() {
  read(n);
  read(k);
  for (int i = (1); i <= (n); i++) read(nex[i]);
  for (int i = (1); i <= (n); i++)
    if (!vis[i]) {
      vis[i] = true;
      c[++cnt] = 1;
      int x = nex[i];
      while (x != i) vis[x] = true, x = nex[x], c[cnt]++;
      cc[c[cnt]]++;
    }
  sort(c + 1, c + 1 + cnt);
  int tmp = k, ansm = 0, ansx = 0;
  f[0] = 1;
  for (int i = (1); i <= (n); i++) {
    if (!cc[i]) continue;
    int pw = 1;
    for (; pw <= cc[i]; cc[i] -= pw, pw <<= 1) f |= f << (i * pw);
    f |= f << (i * cc[i]);
  }
  if (f[k] == 1)
    ansm = k;
  else
    ansm = k + 1;
  tmp = k;
  for (int i = (1); i <= (cnt); i++)
    if (tmp) {
      int t = c[i] / 2;
      if (tmp >= t)
        tmp -= t, ansx += 2 * t, c[i] -= 2 * t;
      else {
        ansx += 2 * tmp;
        c[i] -= tmp;
        tmp = 0;
        break;
      }
    }
  if (!tmp) {
    write(ansm, ' ');
    write(ansx, '\n');
    return 0;
  } else {
    for (int i = (1); i <= (cnt); i++)
      if (c[i] && tmp) tmp--, ansx++;
    write(ansm, ' ');
    write(ansx, '\n');
  }
}

### Prompt

Please provide a CPP coded solution 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;
template <class T>
inline void read(T& num) {
  num = 0;
  bool f = true;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = false;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    num = num * 10 + ch - '0';
    ch = getchar();
  }
  num = f ? num : -num;
}
int out[100];
template <class T>
inline void write(T x, char ch) {
  if (x == 0) {
    putchar('0');
    putchar(ch);
    return;
  }
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  int num = 0;
  while (x) {
    out[num++] = (x % 10);
    x = x / 10;
  }
  for (int i = (num - 1); i >= (0); i--) putchar(out[i] + '0');
  putchar(ch);
}
const double eps = 1e-8;
bitset<1000010> f;
bool vis[1000005];
int c[1000005], cc[1000005], nex[1000005];
int n, k, cnt;
int main() {
  read(n);
  read(k);
  for (int i = (1); i <= (n); i++) read(nex[i]);
  for (int i = (1); i <= (n); i++)
    if (!vis[i]) {
      vis[i] = true;
      c[++cnt] = 1;
      int x = nex[i];
      while (x != i) vis[x] = true, x = nex[x], c[cnt]++;
      cc[c[cnt]]++;
    }
  sort(c + 1, c + 1 + cnt);
  int tmp = k, ansm = 0, ansx = 0;
  f[0] = 1;
  for (int i = (1); i <= (n); i++) {
    if (!cc[i]) continue;
    int pw = 1;
    for (; pw <= cc[i]; cc[i] -= pw, pw <<= 1) f |= f << (i * pw);
    f |= f << (i * cc[i]);
  }
  if (f[k] == 1)
    ansm = k;
  else
    ansm = k + 1;
  tmp = k;
  for (int i = (1); i <= (cnt); i++)
    if (tmp) {
      int t = c[i] / 2;
      if (tmp >= t)
        tmp -= t, ansx += 2 * t, c[i] -= 2 * t;
      else {
        ansx += 2 * tmp;
        c[i] -= tmp;
        tmp = 0;
        break;
      }
    }
  if (!tmp) {
    write(ansm, ' ');
    write(ansx, '\n');
    return 0;
  } else {
    for (int i = (1); i <= (cnt); i++)
      if (c[i] && tmp) tmp--, ansx++;
    write(ansm, ' ');
    write(ansx, '\n');
  }
}

```