#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const long long LINF = INF * (long long)INF;
const int MAX = 1000006;
int p[MAX];
int n, k;
bool us[MAX];
int cnt = 0;
void dfs(int v) {
  while (!us[v]) {
    us[v] = true;
    cnt++;
    v = p[v];
  }
}
map<int, int> cm;
const int MAGIC = 300;
vector<int> v;
bool dp[MAX];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = (0); i < (n); i++) {
    cin >> p[i];
    --p[i];
  }
  vector<int> w;
  for (int i = (0); i < (n); i++) {
    if (!us[i]) {
      cnt = 0;
      dfs(i);
      cm[cnt]++;
      w.push_back(cnt);
    }
  }
  for (__typeof(cm.begin()) it = cm.begin(); it != cm.end(); it++) {
    int x = it->second;
    int val = it->first;
    int lg = 1;
    while (lg * 2 - 1 <= x) {
      v.push_back(lg * val);
      lg *= 2;
    }
    int rem = x - (lg - 1);
    if (rem) v.push_back(rem * val);
  }
  int dpk = min(k, n - k);
  vector<int> nv;
  dp[0] = true;
  for (int i = (0); i < ((int)v.size()); i++) {
    if (dp[dpk]) break;
    int x = v[i];
    for (int j = (k - x + 1) - 1; j >= (0); j--) {
      dp[j + x] |= dp[j];
    }
  }
  int mx = 0;
  int mn = 0;
  if (dp[dpk]) {
    mn = k;
  } else
    mn = k + 1;
  v = w;
  for (int i = (0); i < ((int)v.size()); i++) {
    int take = min(v[i] / 2, k);
    k -= take;
    mx += take * 2;
  }
  for (int i = (0); i < ((int)v.size()); i++)
    if (v[i] % 2 == 1) {
      if (k == 0) break;
      int take = 1;
      k -= take;
      mx += 1;
    }
  cout << mn << " " << mx;
  return 0;
}

### Prompt

Develop a solution in CPP 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;
const double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const long long LINF = INF * (long long)INF;
const int MAX = 1000006;
int p[MAX];
int n, k;
bool us[MAX];
int cnt = 0;
void dfs(int v) {
  while (!us[v]) {
    us[v] = true;
    cnt++;
    v = p[v];
  }
}
map<int, int> cm;
const int MAGIC = 300;
vector<int> v;
bool dp[MAX];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = (0); i < (n); i++) {
    cin >> p[i];
    --p[i];
  }
  vector<int> w;
  for (int i = (0); i < (n); i++) {
    if (!us[i]) {
      cnt = 0;
      dfs(i);
      cm[cnt]++;
      w.push_back(cnt);
    }
  }
  for (__typeof(cm.begin()) it = cm.begin(); it != cm.end(); it++) {
    int x = it->second;
    int val = it->first;
    int lg = 1;
    while (lg * 2 - 1 <= x) {
      v.push_back(lg * val);
      lg *= 2;
    }
    int rem = x - (lg - 1);
    if (rem) v.push_back(rem * val);
  }
  int dpk = min(k, n - k);
  vector<int> nv;
  dp[0] = true;
  for (int i = (0); i < ((int)v.size()); i++) {
    if (dp[dpk]) break;
    int x = v[i];
    for (int j = (k - x + 1) - 1; j >= (0); j--) {
      dp[j + x] |= dp[j];
    }
  }
  int mx = 0;
  int mn = 0;
  if (dp[dpk]) {
    mn = k;
  } else
    mn = k + 1;
  v = w;
  for (int i = (0); i < ((int)v.size()); i++) {
    int take = min(v[i] / 2, k);
    k -= take;
    mx += take * 2;
  }
  for (int i = (0); i < ((int)v.size()); i++)
    if (v[i] % 2 == 1) {
      if (k == 0) break;
      int take = 1;
      k -= take;
      mx += 1;
    }
  cout << mn << " " << mx;
  return 0;
}

```