#include <bits/stdc++.h>
using namespace std;
long long ways(int a, int b, int sum) {
  long long k = sum - a - b;
  if (k < 0) return 0;
  return (k + 2) * (k + 1) / 2;
}
int main() {
  long long N, x, y, c;
  cin >> N >> x >> y >> c;
  x--;
  y--;
  long long left = 0, right = 1e9 + 5, best = 1e9 + 5;
  while (left <= right) {
    long long val = left + (right - left) / 2;
    long long res = 0;
    res += ways(0, 0, val) - ways(N - x, 0, val) - ways(0, N - y, val) +
           ways(N - x, N - y, val);
    res += ways(0, 0, val) - ways(x + 1, 0, val) - ways(0, N - y, val) +
           ways(x + 1, N - y, val);
    res += ways(0, 0, val) - ways(N - x, 0, val) - ways(0, y + 1, val) +
           ways(N - x, y + 1, val);
    res += ways(0, 0, val) - ways(x + 1, 0, val) - ways(0, y + 1, val) +
           ways(x + 1, y + 1, val);
    res -= ways(0, 0, val) - ways(1, 0, val) - ways(0, N - y, val) +
           ways(1, N - y, val);
    res -= ways(0, 0, val) - ways(1, 0, val) - ways(0, N - x, val) +
           ways(1, N - x, val);
    res -= ways(0, 0, val) - ways(1, 0, val) - ways(0, x + 1, val) +
           ways(1, x + 1, val);
    res -= ways(0, 0, val) - ways(1, 0, val) - ways(0, y + 1, val) +
           ways(1, y + 1, val);
    res += 1;
    if (res < c) {
      left = val + 1;
      continue;
    }
    best = min(best, val);
    right = val - 1;
  }
  cout << best << endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 ways(int a, int b, int sum) {
  long long k = sum - a - b;
  if (k < 0) return 0;
  return (k + 2) * (k + 1) / 2;
}
int main() {
  long long N, x, y, c;
  cin >> N >> x >> y >> c;
  x--;
  y--;
  long long left = 0, right = 1e9 + 5, best = 1e9 + 5;
  while (left <= right) {
    long long val = left + (right - left) / 2;
    long long res = 0;
    res += ways(0, 0, val) - ways(N - x, 0, val) - ways(0, N - y, val) +
           ways(N - x, N - y, val);
    res += ways(0, 0, val) - ways(x + 1, 0, val) - ways(0, N - y, val) +
           ways(x + 1, N - y, val);
    res += ways(0, 0, val) - ways(N - x, 0, val) - ways(0, y + 1, val) +
           ways(N - x, y + 1, val);
    res += ways(0, 0, val) - ways(x + 1, 0, val) - ways(0, y + 1, val) +
           ways(x + 1, y + 1, val);
    res -= ways(0, 0, val) - ways(1, 0, val) - ways(0, N - y, val) +
           ways(1, N - y, val);
    res -= ways(0, 0, val) - ways(1, 0, val) - ways(0, N - x, val) +
           ways(1, N - x, val);
    res -= ways(0, 0, val) - ways(1, 0, val) - ways(0, x + 1, val) +
           ways(1, x + 1, val);
    res -= ways(0, 0, val) - ways(1, 0, val) - ways(0, y + 1, val) +
           ways(1, y + 1, val);
    res += 1;
    if (res < c) {
      left = val + 1;
      continue;
    }
    best = min(best, val);
    right = val - 1;
  }
  cout << best << endl;
  return 0;
}

```