#include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9;
int main() {
  int n, x, y, c;
  scanf("%d%d%d%d", &n, &x, &y, &c);
  int lb = 0, ub = n * 2 + 10;
  if (c == 1) ub = 0;
  while (ub - lb > 1) {
    long long mid = (lb + ub) / 2;
    long long sum = 0;
    long long up = max<long long>(mid + 1 - x, 0);
    long long left = max<long long>(mid + 1 - y, 0);
    long long down = max<long long>(mid + x - n, 0);
    long long right = max<long long>(mid + y - n, 0);
    long long cut1 = max<long long>(up - y, 0);
    long long cut2 = max<long long>(up + y - 1 - n, 0);
    long long cut3 = max<long long>(down - y, 0);
    long long cut4 = max<long long>(down + y - 1 - n, 0);
    sum += 2 * mid * mid + 2 * mid + 1;
    sum -= up * up;
    sum -= down * down;
    sum -= left * left;
    sum -= right * right;
    sum += cut1 * (cut1 + 1) / 2;
    sum += cut2 * (cut2 + 1) / 2;
    sum += cut3 * (cut3 + 1) / 2;
    sum += cut4 * (cut4 + 1) / 2;
    if (sum < c)
      lb = mid;
    else
      ub = mid;
  }
  int ans = ub;
  printf("%d\n", ans);
  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;
const int inf = (int)1e9;
int main() {
  int n, x, y, c;
  scanf("%d%d%d%d", &n, &x, &y, &c);
  int lb = 0, ub = n * 2 + 10;
  if (c == 1) ub = 0;
  while (ub - lb > 1) {
    long long mid = (lb + ub) / 2;
    long long sum = 0;
    long long up = max<long long>(mid + 1 - x, 0);
    long long left = max<long long>(mid + 1 - y, 0);
    long long down = max<long long>(mid + x - n, 0);
    long long right = max<long long>(mid + y - n, 0);
    long long cut1 = max<long long>(up - y, 0);
    long long cut2 = max<long long>(up + y - 1 - n, 0);
    long long cut3 = max<long long>(down - y, 0);
    long long cut4 = max<long long>(down + y - 1 - n, 0);
    sum += 2 * mid * mid + 2 * mid + 1;
    sum -= up * up;
    sum -= down * down;
    sum -= left * left;
    sum -= right * right;
    sum += cut1 * (cut1 + 1) / 2;
    sum += cut2 * (cut2 + 1) / 2;
    sum += cut3 * (cut3 + 1) / 2;
    sum += cut4 * (cut4 + 1) / 2;
    if (sum < c)
      lb = mid;
    else
      ub = mid;
  }
  int ans = ub;
  printf("%d\n", ans);
  return 0;
}

```