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 n, x, y, c;
long long calc(long long t) {
  long long int ret = 0;
  ret += 2 * t * (t + 1) - 4 * t + 1;
  if (t > x) {
    ret -= (t - x) * (t - x);
  }
  if (t > n - x + 1) {
    ret -= (t - n + x - 1) * (t - n + x - 1);
  }
  if (t > y) {
    ret -= (t - y) * (t - y);
  }
  if (t > n - y + 1) {
    ret -= (t - n + y - 1) * (t - n + y - 1);
  }
  if (t > x + y) {
    ret += ((t - x - y) * (t - x - y + 1) / 2);
  }
  if (t > x + n - y + 1) {
    ret += ((t - x - n + y - 1) * (t - x - n + y - 1 + 1) / 2);
  }
  if (t > n - x + 1 + y) {
    ret += ((t - n + x - 1 - y) * (t - n + x - 1 - y + 1) / 2);
  }
  if (t > n - x + 1 + n - y + 1) {
    ret += ((t - n + x - 1 - n + y - 1) * (t - n + x - 1 - n + y - 1 + 1) / 2);
  }
  return ret;
}
int main() {
  cin >> n >> x >> y >> c;
  int l = 0, r = 1000000002;
  while (r - l > 1) {
    if (calc((r + l) / 2) < c) {
      l = (r + l) / 2;
    } else {
      r = (r + l) / 2;
    }
  }
  cout << r - 1 << endl;
}
