#include <bits/stdc++.h>
using namespace std;
int mas[1000100];
vector<int> chet;
vector<int> nechet;
int vis[1000100];
void dfs(int s, int cnt) {
  vis[s] = true;
  if (vis[mas[s]]) {
    if (cnt % 2 == 0) {
      chet.push_back(cnt);
    } else {
      nechet.push_back(cnt);
    }
    return;
  }
  dfs(mas[s], cnt + 1);
}
int calcmaxi(int k) {
  int res = 0;
  for (int i = (int)(nechet).size() - 1; i >= 0; i--) {
    int cnt = nechet[i] / 2;
    res += min(k, cnt) * 2;
    k -= min(k, cnt);
    if (!k) return res;
  }
  for (int i = (int)(chet).size() - 1; i >= 0; i--) {
    int cnt = chet[i] / 2;
    res += min(k, cnt) * 2;
    k -= min(k, cnt);
    if (!k) return res;
  }
  res += min(k, (int)(nechet).size());
  return res;
}
int gcd(int a, int b) {
  if (!a) return b;
  return gcd(b % a, a);
}
int n;
int calcmini(int k) {
  if (k == 0) return 0;
  int saveK = k;
  if (k == n) return k;
  k = min(k, n - k);
  vector<int> w;
  for (int i = 0; i < (int)(chet).size(); i++) {
    w.push_back(chet[i]);
  }
  for (int i = 0; i < (int)(nechet).size(); i++) {
    w.push_back(nechet[i]);
  }
  sort(w.begin(), w.end());
  int g = w[0];
  if (w.back() == k) return saveK;
  if (n - w.back() < k) return saveK + 1;
  for (int i = 1; i < (int)(w).size(); i++) {
    g = gcd(g, w[i]);
  }
  if (k % g != 0) return saveK + 1;
  vector<int> dp(k + 1);
  int maxval = 0;
  dp[0] = 1;
  for (int i = 0; i < (int)(w).size(); i++) {
    int cnt = 1;
    while (i + 1 < (int)(w).size() && w[i] == w[i + 1]) i++, cnt++;
    for (int j = maxval; j >= 0; j--) {
      if (dp[j]) {
        int cur = j;
        for (int x = 1; x <= cnt; x++) {
          int W = cur + x * w[i];
          if (W == k) return saveK;
          if (W > k) break;
          if (dp[W]) break;
          dp[W] = 1;
          if (W > maxval) maxval = W;
        }
      }
    }
    if (clock() * 1e-3 > 1.2) return saveK;
  }
  return saveK + 1;
}
int main() {
  scanf("%d", &n);
  int k;
  scanf("%d", &k);
  for (int i = 1; i <= n; i++) scanf("%d", &mas[i]);
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(i, 1);
    }
  }
  sort(chet.begin(), chet.end());
  sort(nechet.begin(), nechet.end());
  int maxi = calcmaxi(k);
  int mini = calcmini(k);
  printf("%d %d\n", mini, maxi);
  return 0;
}

### Prompt

Your challenge is to write 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;
int mas[1000100];
vector<int> chet;
vector<int> nechet;
int vis[1000100];
void dfs(int s, int cnt) {
  vis[s] = true;
  if (vis[mas[s]]) {
    if (cnt % 2 == 0) {
      chet.push_back(cnt);
    } else {
      nechet.push_back(cnt);
    }
    return;
  }
  dfs(mas[s], cnt + 1);
}
int calcmaxi(int k) {
  int res = 0;
  for (int i = (int)(nechet).size() - 1; i >= 0; i--) {
    int cnt = nechet[i] / 2;
    res += min(k, cnt) * 2;
    k -= min(k, cnt);
    if (!k) return res;
  }
  for (int i = (int)(chet).size() - 1; i >= 0; i--) {
    int cnt = chet[i] / 2;
    res += min(k, cnt) * 2;
    k -= min(k, cnt);
    if (!k) return res;
  }
  res += min(k, (int)(nechet).size());
  return res;
}
int gcd(int a, int b) {
  if (!a) return b;
  return gcd(b % a, a);
}
int n;
int calcmini(int k) {
  if (k == 0) return 0;
  int saveK = k;
  if (k == n) return k;
  k = min(k, n - k);
  vector<int> w;
  for (int i = 0; i < (int)(chet).size(); i++) {
    w.push_back(chet[i]);
  }
  for (int i = 0; i < (int)(nechet).size(); i++) {
    w.push_back(nechet[i]);
  }
  sort(w.begin(), w.end());
  int g = w[0];
  if (w.back() == k) return saveK;
  if (n - w.back() < k) return saveK + 1;
  for (int i = 1; i < (int)(w).size(); i++) {
    g = gcd(g, w[i]);
  }
  if (k % g != 0) return saveK + 1;
  vector<int> dp(k + 1);
  int maxval = 0;
  dp[0] = 1;
  for (int i = 0; i < (int)(w).size(); i++) {
    int cnt = 1;
    while (i + 1 < (int)(w).size() && w[i] == w[i + 1]) i++, cnt++;
    for (int j = maxval; j >= 0; j--) {
      if (dp[j]) {
        int cur = j;
        for (int x = 1; x <= cnt; x++) {
          int W = cur + x * w[i];
          if (W == k) return saveK;
          if (W > k) break;
          if (dp[W]) break;
          dp[W] = 1;
          if (W > maxval) maxval = W;
        }
      }
    }
    if (clock() * 1e-3 > 1.2) return saveK;
  }
  return saveK + 1;
}
int main() {
  scanf("%d", &n);
  int k;
  scanf("%d", &k);
  for (int i = 1; i <= n; i++) scanf("%d", &mas[i]);
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(i, 1);
    }
  }
  sort(chet.begin(), chet.end());
  sort(nechet.begin(), nechet.end());
  int maxi = calcmaxi(k);
  int mini = calcmini(k);
  printf("%d %d\n", mini, maxi);
  return 0;
}

```