#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
int n, k, vis[maxn], a[maxn], b[maxn], c[maxn], d[maxn];
bool dp[maxn];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int ans1 = k, ans2 = 0, cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    int number = 0, p = i;
    while (!vis[p]) {
      vis[p] = 1;
      number++;
      p = a[p];
    }
    b[++cnt] = number;
  }
  sort(b + 1, b + cnt + 1);
  int kk = k, num = 0;
  for (int i = 1; i <= cnt; ++i) {
    if (b[i] / 2 <= kk)
      ans2 += b[i] / 2 * 2, kk -= b[i] / 2;
    else
      ans2 += 2 * kk, kk = 0;
    if (b[i] & 1) num++;
  }
  ans2 += min(kk, num);
  num = 0;
  int number = 0;
  for (int i = 1; i <= cnt; ++i) {
    if (b[i] == b[i - 1])
      d[num]++;
    else
      d[++num] = 1, a[num] = b[i];
  }
  cnt = 0;
  for (int i = 1; i <= num; ++i) {
    for (int j = 1; d[i]; j <<= 1) {
      int tep = min(d[i], j);
      c[++cnt] = a[i] * tep;
      d[i] -= tep;
    }
  }
  dp[0] = 1;
  for (int i = 1; i <= cnt; ++i)
    for (int j = k; j >= c[i]; --j) dp[j] |= dp[j - c[i]];
  ans1 += (!dp[k]);
  printf("%d %d\n", ans1, ans2);
  return 0;
}

### 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;
const int maxn = 1e6 + 5;
int n, k, vis[maxn], a[maxn], b[maxn], c[maxn], d[maxn];
bool dp[maxn];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int ans1 = k, ans2 = 0, cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    int number = 0, p = i;
    while (!vis[p]) {
      vis[p] = 1;
      number++;
      p = a[p];
    }
    b[++cnt] = number;
  }
  sort(b + 1, b + cnt + 1);
  int kk = k, num = 0;
  for (int i = 1; i <= cnt; ++i) {
    if (b[i] / 2 <= kk)
      ans2 += b[i] / 2 * 2, kk -= b[i] / 2;
    else
      ans2 += 2 * kk, kk = 0;
    if (b[i] & 1) num++;
  }
  ans2 += min(kk, num);
  num = 0;
  int number = 0;
  for (int i = 1; i <= cnt; ++i) {
    if (b[i] == b[i - 1])
      d[num]++;
    else
      d[++num] = 1, a[num] = b[i];
  }
  cnt = 0;
  for (int i = 1; i <= num; ++i) {
    for (int j = 1; d[i]; j <<= 1) {
      int tep = min(d[i], j);
      c[++cnt] = a[i] * tep;
      d[i] -= tep;
    }
  }
  dp[0] = 1;
  for (int i = 1; i <= cnt; ++i)
    for (int j = k; j >= c[i]; --j) dp[j] |= dp[j - c[i]];
  ans1 += (!dp[k]);
  printf("%d %d\n", ans1, ans2);
  return 0;
}

```