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>
int main() {
  int n, x, y, c;
  int t, tt;
  int t_l, t_r, t_t, t_b, t_rt, t_rb, t_lt, t_lb;
  int cnt;
  scanf("%d%d%d%d", &n, &x, &y, &c);
  t_l = x - 1 + 1;
  t_r = n - x + 1;
  t_t = y - 1 + 1;
  t_b = n - y + 1;
  t_lt = t_l + t_t;
  t_lb = t_l + t_b;
  t_rt = t_r + t_t;
  t_rb = t_r + t_b;
  for (t = 1;; t++) {
    cnt = 2 * t * t - 2 * t + 1;
    if (t > t_l) {
      tt = t - t_l;
      cnt -= tt * tt;
    }
    if (t > t_r) {
      tt = t - t_r;
      cnt -= tt * tt;
    }
    if (t > t_b) {
      tt = t - t_b;
      cnt -= tt * tt;
    }
    if (t > t_t) {
      tt = t - t_t;
      cnt -= tt * tt;
    }
    if (t > t_lt) {
      tt = t - t_lt;
      cnt += tt * (tt + 1) / 2;
    }
    if (t > t_lb) {
      tt = t - t_lb;
      cnt += tt * (tt + 1) / 2;
    }
    if (t > t_rt) {
      tt = t - t_rt;
      cnt += tt * (tt + 1) / 2;
    }
    if (t > t_rb) {
      tt = t - t_rb;
      cnt += tt * (tt + 1) / 2;
    }
    if (cnt >= c) {
      printf("%d\n", t - 1);
      break;
    }
  }
  return 0;
}
