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>
using namespace std;
int main() {
  int now = 1, tt = 0, n, x, y, c;
  scanf("%d%d%d%d", &n, &x, &y, &c);
  while (now < c) {
    tt++;
    int l = min(x - 1, tt);
    now += max(l - max((y + tt - n), 0) + 1, 0) +
           max(l - max((tt - y + 1), 0) + 1, 0) - (l == tt);
    l = min(n - x, tt);
    now += max(l - max((y + tt - n), 1) + 1, 0) +
           max(l - max((tt - y + 1), 1) + 1, 0) - (l == tt);
  }
  printf("%d\n", tt);
  return 0;
}
