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>.
#include <bits/stdc++.h>
long long n, x, y, c;
long long aps(long long n) { return (n * (n + 1)) / 2; }
long long max(long long a, long long b) { return a > b ? a : b; }
long long res(long long t) {
  long long s = t * t + (t + 1) * (t + 1);
  long long top = max(t - y + 1, 0);
  long long bottom = max(y + t - n, 0);
  long long left = max(t - x + 1, 0);
  long long right = max(x + t - n, 0);
  s -= top * top + bottom * bottom + left * left + right * right;
  if (top + left > t + 1) s += aps(top + left - t - 1);
  if (left + bottom > t + 1) s += aps(left + bottom - t - 1);
  if (bottom + right > t + 1) s += aps(bottom + right - t - 1);
  if (right + top > t + 1) s += aps(right + top - t - 1);
  return s;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c);
  long long l = 0, r = c;
  while (l + 1 < r) {
    int m = l + (r - l) * 0.5;
    if (res(m) >= c)
      r = m;
    else
      l = m;
  }
  printf("%lld", res(l) == c ? l : r);
  return 0;
}
