#include <bits/stdc++.h>
long long n, x, y, c;
long long gao(long long mid) {
  long long x1, y1, ans, t, dx, dy;
  ans = (2 * mid - 1) * (2 * mid - 1) - 2 * mid * (mid - 1);
  x1 = x + mid - 1;
  if (x1 > n) {
    t = x1 - n;
    ans -= t * t;
  }
  x1 = x - mid + 1;
  if (x1 < 1) {
    t = 1 - x1;
    ans -= t * t;
  }
  y1 = y + mid - 1;
  if (y1 > n) {
    t = y1 - n;
    ans -= t * t;
  }
  y1 = y - mid + 1;
  if (y1 < 1) {
    t = 1 - y1;
    ans -= t * t;
  }
  dx = n - x + 1;
  dy = n - y + 1;
  if (dx + dy <= mid - 1) {
    t = mid - dx - dy;
    ans += t * (t + 1) / 2;
  }
  dx = n - x + 1;
  dy = y;
  if (dx + dy <= mid - 1) {
    t = mid - dx - dy;
    ans += t * (t + 1) / 2;
  }
  dx = x;
  dy = n - y + 1;
  if (dx + dy <= mid - 1) {
    t = mid - dx - dy;
    ans += t * (t + 1) / 2;
  }
  dx = x;
  dy = y;
  if (dx + dy <= mid - 1) {
    t = mid - dx - dy;
    ans += t * (t + 1) / 2;
  }
  return ans;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c);
  long long l, r, mid;
  l = 1;
  r = 2 * n;
  while (l <= r) {
    mid = (l + r) / 2;
    if (gao(mid) >= c)
      r = mid - 1;
    else
      l = mid + 1;
  }
  printf("%I64d\n", r);
}

### Prompt

Develop a solution in Cpp 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>
long long n, x, y, c;
long long gao(long long mid) {
  long long x1, y1, ans, t, dx, dy;
  ans = (2 * mid - 1) * (2 * mid - 1) - 2 * mid * (mid - 1);
  x1 = x + mid - 1;
  if (x1 > n) {
    t = x1 - n;
    ans -= t * t;
  }
  x1 = x - mid + 1;
  if (x1 < 1) {
    t = 1 - x1;
    ans -= t * t;
  }
  y1 = y + mid - 1;
  if (y1 > n) {
    t = y1 - n;
    ans -= t * t;
  }
  y1 = y - mid + 1;
  if (y1 < 1) {
    t = 1 - y1;
    ans -= t * t;
  }
  dx = n - x + 1;
  dy = n - y + 1;
  if (dx + dy <= mid - 1) {
    t = mid - dx - dy;
    ans += t * (t + 1) / 2;
  }
  dx = n - x + 1;
  dy = y;
  if (dx + dy <= mid - 1) {
    t = mid - dx - dy;
    ans += t * (t + 1) / 2;
  }
  dx = x;
  dy = n - y + 1;
  if (dx + dy <= mid - 1) {
    t = mid - dx - dy;
    ans += t * (t + 1) / 2;
  }
  dx = x;
  dy = y;
  if (dx + dy <= mid - 1) {
    t = mid - dx - dy;
    ans += t * (t + 1) / 2;
  }
  return ans;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c);
  long long l, r, mid;
  l = 1;
  r = 2 * n;
  while (l <= r) {
    mid = (l + r) / 2;
    if (gao(mid) >= c)
      r = mid - 1;
    else
      l = mid + 1;
  }
  printf("%I64d\n", r);
}

```