#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, x, y, c;
  scanf("%d%d%d%d", &n, &x, &y, &c);
  long long num = 1;
  long long tnum = 1;
  long long s = 0;
  int cnt = 0;
  long long t;
  while (tnum < c) {
    cnt++;
    s += 4;
    num += s;
    tnum = num;
    t = x - cnt;
    if (t < 1) tnum -= (1 - t) * (1 - t);
    t = x + cnt;
    if (t > n) tnum -= (t - n) * (t - n);
    t = y - cnt;
    if (t < 1) tnum -= (1 - t) * (1 - t);
    t = y + cnt;
    if (t > n) tnum -= (t - n) * (t - n);
    t = (y + cnt - n) - x;
    if (t >= 1) tnum += (1 + t) * t / 2;
    t = (1 - y + cnt) - x;
    if (t >= 1) tnum += (1 + t) * t / 2;
    t = 1 - (y - cnt) + x - n - 1;
    if (t >= 1) tnum += (1 + t) * t / 2;
    t = y + cnt - n + x - n - 1;
    if (t >= 1) tnum += (1 + t) * t / 2;
  }
  printf("%d\n", cnt);
}

### Prompt

Your task is to create 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;
int main() {
  int n, x, y, c;
  scanf("%d%d%d%d", &n, &x, &y, &c);
  long long num = 1;
  long long tnum = 1;
  long long s = 0;
  int cnt = 0;
  long long t;
  while (tnum < c) {
    cnt++;
    s += 4;
    num += s;
    tnum = num;
    t = x - cnt;
    if (t < 1) tnum -= (1 - t) * (1 - t);
    t = x + cnt;
    if (t > n) tnum -= (t - n) * (t - n);
    t = y - cnt;
    if (t < 1) tnum -= (1 - t) * (1 - t);
    t = y + cnt;
    if (t > n) tnum -= (t - n) * (t - n);
    t = (y + cnt - n) - x;
    if (t >= 1) tnum += (1 + t) * t / 2;
    t = (1 - y + cnt) - x;
    if (t >= 1) tnum += (1 + t) * t / 2;
    t = 1 - (y - cnt) + x - n - 1;
    if (t >= 1) tnum += (1 + t) * t / 2;
    t = y + cnt - n + x - n - 1;
    if (t >= 1) tnum += (1 + t) * t / 2;
  }
  printf("%d\n", cnt);
}

```