#include <bits/stdc++.h>
using namespace std;
const int INF = (int)1e9 + 123;
const int N = (int)1e6 + 123;
const int B = 170;
int n, k;
int p[N];
int cnt;
int a[N];
bool used[N];
int q[N];
int need[N];
int res[N];
void run() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
  }
  for (int i = 0; i < n; i++) {
    if (used[i]) {
      continue;
    }
    int v = i;
    while (p[v] != i) {
      used[v] = 1;
      a[cnt]++;
      v = p[v];
    }
    a[cnt]++;
    used[v] = 1;
    cnt++;
  }
  int mi = 0;
  int ma = 0;
  int rk = k;
  int tmp = 0;
  for (int i = 0; i < cnt; i++) {
    tmp += a[i] % 2;
    ma += min(a[i] - a[i] % 2, 2 * rk);
    rk -= min(a[i] / 2, rk);
  }
  if (rk) {
    ma += min(rk, tmp);
  }
  for (int i = 0; i < cnt; i++) {
    q[a[i]]++;
  }
  bitset<N> dp;
  dp[0] = 1;
  for (int i = 1; i < N; i++) {
    if (!q[i]) {
      continue;
    }
    for (int j = 0; j < 20; j++) {
      if ((1 << j) <= q[i]) {
        q[i] -= (1 << j);
        res[(1 << j) * i]++;
      }
    }
    res[q[i] * i]++;
  }
  for (int i = 1; i < N; i++) {
    for (int j = 0; j < res[i]; j++) {
      dp |= dp << i;
    }
  }
  mi = dp[k] ? k : k + 1;
  cout << mi << " " << ma << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  run();
  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;
const int INF = (int)1e9 + 123;
const int N = (int)1e6 + 123;
const int B = 170;
int n, k;
int p[N];
int cnt;
int a[N];
bool used[N];
int q[N];
int need[N];
int res[N];
void run() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
  }
  for (int i = 0; i < n; i++) {
    if (used[i]) {
      continue;
    }
    int v = i;
    while (p[v] != i) {
      used[v] = 1;
      a[cnt]++;
      v = p[v];
    }
    a[cnt]++;
    used[v] = 1;
    cnt++;
  }
  int mi = 0;
  int ma = 0;
  int rk = k;
  int tmp = 0;
  for (int i = 0; i < cnt; i++) {
    tmp += a[i] % 2;
    ma += min(a[i] - a[i] % 2, 2 * rk);
    rk -= min(a[i] / 2, rk);
  }
  if (rk) {
    ma += min(rk, tmp);
  }
  for (int i = 0; i < cnt; i++) {
    q[a[i]]++;
  }
  bitset<N> dp;
  dp[0] = 1;
  for (int i = 1; i < N; i++) {
    if (!q[i]) {
      continue;
    }
    for (int j = 0; j < 20; j++) {
      if ((1 << j) <= q[i]) {
        q[i] -= (1 << j);
        res[(1 << j) * i]++;
      }
    }
    res[q[i] * i]++;
  }
  for (int i = 1; i < N; i++) {
    for (int j = 0; j < res[i]; j++) {
      dp |= dp << i;
    }
  }
  mi = dp[k] ? k : k + 1;
  cout << mi << " " << ma << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  run();
  return 0;
}

```