#include <bits/stdc++.h>
using namespace std;
long long num(long long c) { return c * c; }
long long num2(long long c) { return (c + 1) * c / (long long)2; }
long long cal(int n, int x, int y, long long t) {
  long long blocks = 2 * t * t + 2 * t + 1;
  if (y <= t) blocks -= num(t - (long long)y + 1);
  if (x <= t) blocks -= num(t - (long long)x + 1);
  if (n - y < t) blocks -= num(t - n + (long long)y);
  if (n - x < t) blocks -= num(t - n + (long long)x);
  int tmp;
  tmp = x + y - 2 + 1;
  if (tmp < t) blocks += num2(t - tmp);
  tmp = n - 1 - (y - x) + 1;
  if (tmp < t) blocks += num2(t - tmp);
  tmp = 2 * n - x - y + 1;
  if (t > tmp) blocks += num2(t - tmp);
  tmp = (y - x) - (1 - n) + 1;
  if (tmp < t) blocks += num2(t - tmp);
  return blocks;
}
int main() {
  int n, x, y;
  long long c;
  cin >> n >> x >> y >> c;
  int ans = 0;
  int f, l;
  for (f = 0, l = 2 * n; f <= l;) {
    int m = (f + l) / 2;
    long long blocks = cal(n, x, y, m);
    if (blocks >= c) {
      ans = m;
      l = m - 1;
    } else
      f = m + 1;
  }
  printf("%d", ans);
  return 0;
}

### 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>
using namespace std;
long long num(long long c) { return c * c; }
long long num2(long long c) { return (c + 1) * c / (long long)2; }
long long cal(int n, int x, int y, long long t) {
  long long blocks = 2 * t * t + 2 * t + 1;
  if (y <= t) blocks -= num(t - (long long)y + 1);
  if (x <= t) blocks -= num(t - (long long)x + 1);
  if (n - y < t) blocks -= num(t - n + (long long)y);
  if (n - x < t) blocks -= num(t - n + (long long)x);
  int tmp;
  tmp = x + y - 2 + 1;
  if (tmp < t) blocks += num2(t - tmp);
  tmp = n - 1 - (y - x) + 1;
  if (tmp < t) blocks += num2(t - tmp);
  tmp = 2 * n - x - y + 1;
  if (t > tmp) blocks += num2(t - tmp);
  tmp = (y - x) - (1 - n) + 1;
  if (tmp < t) blocks += num2(t - tmp);
  return blocks;
}
int main() {
  int n, x, y;
  long long c;
  cin >> n >> x >> y >> c;
  int ans = 0;
  int f, l;
  for (f = 0, l = 2 * n; f <= l;) {
    int m = (f + l) / 2;
    long long blocks = cal(n, x, y, m);
    if (blocks >= c) {
      ans = m;
      l = m - 1;
    } else
      f = m + 1;
  }
  printf("%d", ans);
  return 0;
}

```