#include <bits/stdc++.h>
using namespace std;
const int mo = (int)1e9 + 7, inf = (int)1e9;
const long long linf = (long long)1e18;
int n, x, y, c;
long long calc2(long long x) { return x <= 0 ? 0 : (1 + x) * x / 2; }
long long calc1(long long x) { return x <= 0 ? 0 : x * x; }
long long check(long long t) {
  return 1 + (1 + t) * t * 2 - calc1(t - x + 1) - calc1(t - (n - x)) -
         calc1(t - y + 1) - calc1(t - (n - y)) + calc2(t - x - y + 1) +
         calc2(t - (n - x + 1) - y + 1) + calc2(t - x - (n - y + 1) + 1) +
         calc2(t - (n - x + 1) - (n - y + 1) + 1);
}
int main() {
  scanf("%d%d%d%d", &n, &x, &y, &c);
  long long l = -1, r = c;
  while (r - l > 1) {
    long long t = r - (r - l) / 2;
    if (check(t) < c)
      l = t;
    else
      r = t;
  }
  printf("%I64d\n", r);
  scanf("\n");
}

### 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 mo = (int)1e9 + 7, inf = (int)1e9;
const long long linf = (long long)1e18;
int n, x, y, c;
long long calc2(long long x) { return x <= 0 ? 0 : (1 + x) * x / 2; }
long long calc1(long long x) { return x <= 0 ? 0 : x * x; }
long long check(long long t) {
  return 1 + (1 + t) * t * 2 - calc1(t - x + 1) - calc1(t - (n - x)) -
         calc1(t - y + 1) - calc1(t - (n - y)) + calc2(t - x - y + 1) +
         calc2(t - (n - x + 1) - y + 1) + calc2(t - x - (n - y + 1) + 1) +
         calc2(t - (n - x + 1) - (n - y + 1) + 1);
}
int main() {
  scanf("%d%d%d%d", &n, &x, &y, &c);
  long long l = -1, r = c;
  while (r - l > 1) {
    long long t = r - (r - l) / 2;
    if (check(t) < c)
      l = t;
    else
      r = t;
  }
  printf("%I64d\n", r);
  scanf("\n");
}

```