#include <bits/stdc++.h>
using namespace std;
constexpr static int MAXN = 1e6 + 10;
int n, k;
int p[MAXN];
bool vis[MAXN];
bool reach[MAXN];
vector<vector<int>> cycles;
int solve1() {
  unordered_map<int, int> m;
  for (auto& c : cycles) m[c.size()]++;
  reach[0] = true;
  int sum = 0;
  int goal = min(k, n - k);
  for (auto& kv : m) {
    int len = kv.first;
    int cnt = kv.second;
    for (int i = min(goal, sum); i >= 0; i--) {
      if (reach[i]) {
        for (int j = 1; j <= cnt && i + j * len <= goal; j++) {
          if (reach[i + j * len]) break;
          reach[i + j * len] = true;
        }
      }
    }
    sum += cnt * len;
    if (reach[goal]) return k;
  }
  if (reach[goal]) return k;
  return k + 1;
}
int solve2() {
  int odd = 0;
  int total = 0;
  int used = 0;
  for (auto& c : cycles) {
    int len = c.size();
    if (len % 2 == 1) odd++;
    int cur = min(len / 2, k - used);
    used += cur;
    total += 2 * cur;
  }
  total += min(k - used, odd);
  return total;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
    p[i]--;
  }
  for (int i = 0; i < n; i++) {
    if (vis[i]) continue;
    vector<int> c;
    c.push_back(i);
    vis[i] = true;
    int j = i;
    while (p[j] != i) {
      j = p[j];
      c.push_back(j);
      vis[j] = true;
    }
    cycles.push_back(c);
  }
  printf("%d %d\n", solve1(), solve2());
}

### 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;
constexpr static int MAXN = 1e6 + 10;
int n, k;
int p[MAXN];
bool vis[MAXN];
bool reach[MAXN];
vector<vector<int>> cycles;
int solve1() {
  unordered_map<int, int> m;
  for (auto& c : cycles) m[c.size()]++;
  reach[0] = true;
  int sum = 0;
  int goal = min(k, n - k);
  for (auto& kv : m) {
    int len = kv.first;
    int cnt = kv.second;
    for (int i = min(goal, sum); i >= 0; i--) {
      if (reach[i]) {
        for (int j = 1; j <= cnt && i + j * len <= goal; j++) {
          if (reach[i + j * len]) break;
          reach[i + j * len] = true;
        }
      }
    }
    sum += cnt * len;
    if (reach[goal]) return k;
  }
  if (reach[goal]) return k;
  return k + 1;
}
int solve2() {
  int odd = 0;
  int total = 0;
  int used = 0;
  for (auto& c : cycles) {
    int len = c.size();
    if (len % 2 == 1) odd++;
    int cur = min(len / 2, k - used);
    used += cur;
    total += 2 * cur;
  }
  total += min(k - used, odd);
  return total;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
    p[i]--;
  }
  for (int i = 0; i < n; i++) {
    if (vis[i]) continue;
    vector<int> c;
    c.push_back(i);
    vis[i] = true;
    int j = i;
    while (p[j] != i) {
      j = p[j];
      c.push_back(j);
      vis[j] = true;
    }
    cycles.push_back(c);
  }
  printf("%d %d\n", solve1(), solve2());
}

```