#include <bits/stdc++.h>
int n;
long long m;
int xi, yi;
long long check(long long k) {
  long long sum = (2 * k * k - 2 * k + 1);
  if (k > xi + 1) {
    long long tmp = k - (xi + 1);
    sum -= ((2 * tmp * tmp - 2 * tmp + 1) - (2 * tmp - 1)) / 2 + (2 * tmp - 1);
  }
  if (k > yi + 1) {
    long long tmp = k - (yi + 1);
    sum -= ((2 * tmp * tmp - 2 * tmp + 1) - (2 * tmp - 1)) / 2 + (2 * tmp - 1);
  }
  if (k > n - xi) {
    long long tmp = k - (n - xi);
    sum -= ((2 * tmp * tmp - 2 * tmp + 1) - (2 * tmp - 1)) / 2 + (2 * tmp - 1);
  }
  if (k > n - yi) {
    long long tmp = k - (n - yi);
    sum -= ((2 * tmp * tmp - 2 * tmp + 1) - (2 * tmp - 1)) / 2 + (2 * tmp - 1);
  }
  if (k - (xi + 1) > n - yi) {
    long long tmp = (k - (xi + 1)) - (n - yi);
    sum += ((2 * tmp * tmp - 2 * tmp + 1) - tmp - 3 * (tmp - 1)) / 4 + tmp +
           (tmp - 1);
  }
  if (k - (xi + 1) > yi + 1) {
    long long tmp = (k - (xi + 1)) - (yi + 1);
    sum += ((2 * tmp * tmp - 2 * tmp + 1) - tmp - 3 * (tmp - 1)) / 4 + tmp +
           (tmp - 1);
  }
  if (k - (n - xi) > yi + 1) {
    long long tmp = (k - (n - xi)) - (yi + 1);
    sum += ((2 * tmp * tmp - 2 * tmp + 1) - tmp - 3 * (tmp - 1)) / 4 + tmp +
           (tmp - 1);
  }
  if (k - (n - xi) > n - yi) {
    long long tmp = (k - (n - xi)) - (n - yi);
    sum += ((2 * tmp * tmp - 2 * tmp + 1) - tmp - 3 * (tmp - 1)) / 4 + tmp +
           (tmp - 1);
  }
  return sum;
}
int main() {
  scanf("%d %d %d", &n, &xi, &yi);
  xi--;
  yi--;
  scanf("%I64d", &m);
  long long left = 1, right = n + 1;
  right *= 3;
  for (; left < right;) {
    long long mid = (left + right) / 2;
    long long sum = check(mid);
    if (sum >= m)
      right = mid;
    else
      left = mid + 1;
  }
  printf("%I64d\n", left - 1);
}

### Prompt

Generate a CPP solution to the following problem:

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>
int n;
long long m;
int xi, yi;
long long check(long long k) {
  long long sum = (2 * k * k - 2 * k + 1);
  if (k > xi + 1) {
    long long tmp = k - (xi + 1);
    sum -= ((2 * tmp * tmp - 2 * tmp + 1) - (2 * tmp - 1)) / 2 + (2 * tmp - 1);
  }
  if (k > yi + 1) {
    long long tmp = k - (yi + 1);
    sum -= ((2 * tmp * tmp - 2 * tmp + 1) - (2 * tmp - 1)) / 2 + (2 * tmp - 1);
  }
  if (k > n - xi) {
    long long tmp = k - (n - xi);
    sum -= ((2 * tmp * tmp - 2 * tmp + 1) - (2 * tmp - 1)) / 2 + (2 * tmp - 1);
  }
  if (k > n - yi) {
    long long tmp = k - (n - yi);
    sum -= ((2 * tmp * tmp - 2 * tmp + 1) - (2 * tmp - 1)) / 2 + (2 * tmp - 1);
  }
  if (k - (xi + 1) > n - yi) {
    long long tmp = (k - (xi + 1)) - (n - yi);
    sum += ((2 * tmp * tmp - 2 * tmp + 1) - tmp - 3 * (tmp - 1)) / 4 + tmp +
           (tmp - 1);
  }
  if (k - (xi + 1) > yi + 1) {
    long long tmp = (k - (xi + 1)) - (yi + 1);
    sum += ((2 * tmp * tmp - 2 * tmp + 1) - tmp - 3 * (tmp - 1)) / 4 + tmp +
           (tmp - 1);
  }
  if (k - (n - xi) > yi + 1) {
    long long tmp = (k - (n - xi)) - (yi + 1);
    sum += ((2 * tmp * tmp - 2 * tmp + 1) - tmp - 3 * (tmp - 1)) / 4 + tmp +
           (tmp - 1);
  }
  if (k - (n - xi) > n - yi) {
    long long tmp = (k - (n - xi)) - (n - yi);
    sum += ((2 * tmp * tmp - 2 * tmp + 1) - tmp - 3 * (tmp - 1)) / 4 + tmp +
           (tmp - 1);
  }
  return sum;
}
int main() {
  scanf("%d %d %d", &n, &xi, &yi);
  xi--;
  yi--;
  scanf("%I64d", &m);
  long long left = 1, right = n + 1;
  right *= 3;
  for (; left < right;) {
    long long mid = (left + right) / 2;
    long long sum = check(mid);
    if (sum >= m)
      right = mid;
    else
      left = mid + 1;
  }
  printf("%I64d\n", left - 1);
}

```