#include <bits/stdc++.h>
using namespace std;
long long int calculate(long long int x, long long int y, long long int time) {
  long long int ans = 0;
  long long int mini = min(x, y);
  if (time < mini)
    return ((time + 1) * (time + 2)) / 2;
  else if (time >= x + y - 2)
    return (x * y);
  else {
    long long int ans = (mini * (mini + 1) / 2);
    time -= mini - 1;
    if (time <= max(x, y) - mini) {
      ans += time * mini;
    } else {
      time -= max(x, y) - mini;
      ans += (max(x, y) - mini) * mini;
      ans += ((time) * (2 * (mini - 1) + (time - 1) * (-1))) / 2;
    }
    return ans;
  }
}
long long int n, k;
bool cal(long long int x, long long int y, long long int time) {
  long long int ans = 0;
  ans -= min(x, time + 1);
  ans -= min(y, time + 1);
  ans -= min(n - y + 1, time + 1);
  ans -= min(n - x + 1, time + 1);
  ans += calculate(x, y, time);
  ans += calculate(n - x + 1, y, time);
  ans += calculate(x, n - y + 1, time);
  ans += calculate(n - x + 1, n - y + 1, time);
  ans += 1;
  if (ans >= k) return true;
  return false;
}
int main() {
  long long int x, y;
  scanf("%lld", &n);
  scanf("%lld", &x);
  scanf("%lld", &y);
  scanf("%lld", &k);
  long long int low = 0;
  long long int high = 2000000000 * 1ll * 2000000000;
  while (low < high) {
    long long int mid = (low + high) / 2;
    if (cal(x, y, mid))
      high = mid;
    else
      low = mid + 1;
  }
  printf("%lld\n", low);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

Mr. Bender has a digital table of size n × n, each cell can be switched on or off. He wants the field to have at least c switched on squares. When this condition is fulfilled, Mr Bender will be happy.

We'll consider the table rows numbered from top to bottom from 1 to n, and the columns — numbered from left to right from 1 to n. Initially there is exactly one switched on cell with coordinates (x, y) (x is the row number, y is the column number), and all other cells are switched off. Then each second we switch on the cells that are off but have the side-adjacent cells that are on.

For a cell with coordinates (x, y) the side-adjacent cells are cells with coordinates (x - 1, y), (x + 1, y), (x, y - 1), (x, y + 1).

In how many seconds will Mr. Bender get happy?

Input

The first line contains four space-separated integers n, x, y, c (1 ≤ n, c ≤ 109; 1 ≤ x, y ≤ n; c ≤ n2).

Output

In a single line print a single integer — the answer to the problem.

Examples

Input

6 4 3 1


Output

0


Input

9 3 8 10


Output

2

Note

Initially the first test has one painted cell, so the answer is 0. In the second test all events will go as is shown on the figure. <image>.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int calculate(long long int x, long long int y, long long int time) {
  long long int ans = 0;
  long long int mini = min(x, y);
  if (time < mini)
    return ((time + 1) * (time + 2)) / 2;
  else if (time >= x + y - 2)
    return (x * y);
  else {
    long long int ans = (mini * (mini + 1) / 2);
    time -= mini - 1;
    if (time <= max(x, y) - mini) {
      ans += time * mini;
    } else {
      time -= max(x, y) - mini;
      ans += (max(x, y) - mini) * mini;
      ans += ((time) * (2 * (mini - 1) + (time - 1) * (-1))) / 2;
    }
    return ans;
  }
}
long long int n, k;
bool cal(long long int x, long long int y, long long int time) {
  long long int ans = 0;
  ans -= min(x, time + 1);
  ans -= min(y, time + 1);
  ans -= min(n - y + 1, time + 1);
  ans -= min(n - x + 1, time + 1);
  ans += calculate(x, y, time);
  ans += calculate(n - x + 1, y, time);
  ans += calculate(x, n - y + 1, time);
  ans += calculate(n - x + 1, n - y + 1, time);
  ans += 1;
  if (ans >= k) return true;
  return false;
}
int main() {
  long long int x, y;
  scanf("%lld", &n);
  scanf("%lld", &x);
  scanf("%lld", &y);
  scanf("%lld", &k);
  long long int low = 0;
  long long int high = 2000000000 * 1ll * 2000000000;
  while (low < high) {
    long long int mid = (low + high) / 2;
    if (cal(x, y, mid))
      high = mid;
    else
      low = mid + 1;
  }
  printf("%lld\n", low);
  return 0;
}

```