#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long _gets(long long step) {
  long long s = (step + 1) * (step + 1) + step * step;
  long long left = 0, right = 0, top = 0, bottom = 0;
  long long lb = 0, rb = 0, lt = 0, rt = 0;
  if (step - x > 0) {
    left = (step - x) * (step - x);
    long long h = step - x - y - 1;
    if (h > 0) lb = ((1 + h) * h) / 2;
    h = step - x - 1 - (n - 1 - y);
    if (h > 0) lt = ((1 + h) * h) / 2;
  }
  if (step - y > 0) bottom = (step - y) * (step - y);
  if (step - (n - 1 - x) > 0) {
    right = (step - (n - 1 - x)) * (step - (n - 1 - x));
    long long h = (step - (n - 1 - x)) - y - 1;
    if (h > 0) rb = ((1 + h) * h) / 2;
    h = (step - (n - 1 - x)) - 1 - (n - 1 - y);
    if (h > 0) rt = ((1 + h) * h) / 2;
  }
  if (step - (n - 1 - y) > 0) top = (step - (n - 1 - y)) * (step - (n - 1 - y));
  return s - left - right - top - bottom + lb + rb + lt + rt;
}
int main(int argc, char** argv) {
  cin >> n >> x >> y >> c;
  if (c == 1) {
    cout << 0 << endl;
    return 0;
  }
  --x;
  --y;
  long long l = 0;
  long long r = 2 * n + 1;
  while (l + 1 < r) {
    long long m = (l + r) / 2;
    long long s = _gets(m);
    if (s >= c)
      r = m;
    else
      l = m;
  }
  cout << r << endl;
}

### 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;
long long n, x, y, c;
long long _gets(long long step) {
  long long s = (step + 1) * (step + 1) + step * step;
  long long left = 0, right = 0, top = 0, bottom = 0;
  long long lb = 0, rb = 0, lt = 0, rt = 0;
  if (step - x > 0) {
    left = (step - x) * (step - x);
    long long h = step - x - y - 1;
    if (h > 0) lb = ((1 + h) * h) / 2;
    h = step - x - 1 - (n - 1 - y);
    if (h > 0) lt = ((1 + h) * h) / 2;
  }
  if (step - y > 0) bottom = (step - y) * (step - y);
  if (step - (n - 1 - x) > 0) {
    right = (step - (n - 1 - x)) * (step - (n - 1 - x));
    long long h = (step - (n - 1 - x)) - y - 1;
    if (h > 0) rb = ((1 + h) * h) / 2;
    h = (step - (n - 1 - x)) - 1 - (n - 1 - y);
    if (h > 0) rt = ((1 + h) * h) / 2;
  }
  if (step - (n - 1 - y) > 0) top = (step - (n - 1 - y)) * (step - (n - 1 - y));
  return s - left - right - top - bottom + lb + rb + lt + rt;
}
int main(int argc, char** argv) {
  cin >> n >> x >> y >> c;
  if (c == 1) {
    cout << 0 << endl;
    return 0;
  }
  --x;
  --y;
  long long l = 0;
  long long r = 2 * n + 1;
  while (l + 1 < r) {
    long long m = (l + r) / 2;
    long long s = _gets(m);
    if (s >= c)
      r = m;
    else
      l = m;
  }
  cout << r << endl;
}

```