#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void MIN(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline void MAX(T &a, T b) {
  if (a < b) a = b;
}
const int N = 1e6 + 5;
int n, k, p[N];
int sz[N], gp;
bool vis[N];
int cnt[N];
bitset<N> ok;
int main() {
  scanf("%d %d", &n, &k);
  k = n - k;
  for (int i = 1; i <= n; i++) {
    scanf("%d", p + i);
  }
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    int cur = i, amt = 1;
    while (p[cur] != i) {
      cur = p[cur];
      vis[cur] = true;
      amt++;
    }
    sz[gp++] = amt;
  }
  int mn = 0, br = k;
  for (int i = 0; i < gp; i++) {
    br -= sz[i] / 2;
  }
  for (int i = 0; i < gp; i++) {
    if (br <= 0) break;
    if (sz[i] & 1) br--, mn++;
  }
  if (br > 0) mn += br * 2;
  for (int i = 0; i < gp; i++) {
    cnt[sz[i]]++;
  }
  for (int i = 1; i <= n; i++) {
    if (cnt[i] <= 2) continue;
    if (cnt[i] & 1) {
      cnt[2 * i] += (cnt[i] - 1) / 2;
      cnt[i] = 1;
    } else {
      cnt[2 * i] += (cnt[i] - 2) / 2;
      cnt[i] = 2;
    }
  }
  ok[0] = true;
  for (int i = 1; i <= n; i++) {
    while (cnt[i]--) {
      ok |= ok << i;
    }
  }
  int mx = ok[k] ? k : k - 1;
  printf("%d %d\n", n - mx, n - mn);
}

### Prompt

Generate a cpp solution 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;
template <class T>
inline void MIN(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline void MAX(T &a, T b) {
  if (a < b) a = b;
}
const int N = 1e6 + 5;
int n, k, p[N];
int sz[N], gp;
bool vis[N];
int cnt[N];
bitset<N> ok;
int main() {
  scanf("%d %d", &n, &k);
  k = n - k;
  for (int i = 1; i <= n; i++) {
    scanf("%d", p + i);
  }
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    int cur = i, amt = 1;
    while (p[cur] != i) {
      cur = p[cur];
      vis[cur] = true;
      amt++;
    }
    sz[gp++] = amt;
  }
  int mn = 0, br = k;
  for (int i = 0; i < gp; i++) {
    br -= sz[i] / 2;
  }
  for (int i = 0; i < gp; i++) {
    if (br <= 0) break;
    if (sz[i] & 1) br--, mn++;
  }
  if (br > 0) mn += br * 2;
  for (int i = 0; i < gp; i++) {
    cnt[sz[i]]++;
  }
  for (int i = 1; i <= n; i++) {
    if (cnt[i] <= 2) continue;
    if (cnt[i] & 1) {
      cnt[2 * i] += (cnt[i] - 1) / 2;
      cnt[i] = 1;
    } else {
      cnt[2 * i] += (cnt[i] - 2) / 2;
      cnt[i] = 2;
    }
  }
  ok[0] = true;
  for (int i = 1; i <= n; i++) {
    while (cnt[i]--) {
      ok |= ok << i;
    }
  }
  int mx = ok[k] ? k : k - 1;
  printf("%d %d\n", n - mx, n - mn);
}

```