#include <bits/stdc++.h>
using namespace std;
long long n, X, Y, C;
long long cal(long long x, long long y, long long t) {
  if (x > y) swap(x, y);
  if (t <= x) return t * (t + 1) / 2;
  if (t <= y) return x * (t - x) + x * (x + 1) / 2;
  if (t <= x + y - 1)
    return x * (t - x) + x * (x + 1) / 2 - (t - y) * (t - y + 1) / 2;
  return x * y;
}
int main() {
  long long l, r, m;
  scanf("%I64d%I64d%I64d%I64d", &n, &X, &Y, &C);
  l = 0, r = 1LL << 32;
  while (l < r) {
    m = (l + r) >> 1;
    long long s = 1 + cal(1, X - 1, m) + cal(1, n - X, m) + cal(1, Y - 1, m) +
                  cal(1, n - Y, m) + cal(X - 1, Y - 1, m - 1) +
                  cal(X - 1, n - Y, m - 1) + cal(n - X, Y - 1, m - 1) +
                  cal(n - X, n - Y, m - 1);
    if (s >= C)
      r = m;
    else
      l = m + 1;
  }
  cout << l << endl;
  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 n, X, Y, C;
long long cal(long long x, long long y, long long t) {
  if (x > y) swap(x, y);
  if (t <= x) return t * (t + 1) / 2;
  if (t <= y) return x * (t - x) + x * (x + 1) / 2;
  if (t <= x + y - 1)
    return x * (t - x) + x * (x + 1) / 2 - (t - y) * (t - y + 1) / 2;
  return x * y;
}
int main() {
  long long l, r, m;
  scanf("%I64d%I64d%I64d%I64d", &n, &X, &Y, &C);
  l = 0, r = 1LL << 32;
  while (l < r) {
    m = (l + r) >> 1;
    long long s = 1 + cal(1, X - 1, m) + cal(1, n - X, m) + cal(1, Y - 1, m) +
                  cal(1, n - Y, m) + cal(X - 1, Y - 1, m - 1) +
                  cal(X - 1, n - Y, m - 1) + cal(n - X, Y - 1, m - 1) +
                  cal(n - X, n - Y, m - 1);
    if (s >= C)
      r = m;
    else
      l = m + 1;
  }
  cout << l << endl;
  return 0;
}

```