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;
long long n, x, y, c, t, cont = 1;
int main() {
  scanf("%I64d %I64d %I64d %I64d", &n, &x, &y, &c);
  long long donde[4] = {y - 1, x - 1, n - y, n - x};
  while (cont < c) {
    for (int i = 0; i < 4; i += 2) {
      cont += 1 + min(donde[(i + 1) & 3], t) + min(donde[(i + 3) & 3], t);
      if (t >= donde[i]) cont--;
      cont -= min(donde[(i + 1) & 3], max(t - donde[i], (long long)0));
      cont -= min(donde[(i + 3) & 3], max(t - donde[i], (long long)0));
    }
    if (t < donde[1]) cont++;
    if (t < donde[3]) cont++;
    t++;
  }
  printf("%I64d", t);
}
