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 int fun(long long int n) { return n * n; }
long long int sum(long long int w, long long int h, long long int l) {
  long long int n = w + h - l;
  if (n > 0) return ((1 + n) * n) / 2;
  return 0;
}
long long int cal(long long int n, long long int x, long long int y,
                  long long int t) {
  long long int ans = t * t + (t + 1) * (t + 1);
  long long int u = t - (x - 1);
  long long int d = t - (n - x);
  long long int l = t - (y - 1);
  long long int r = t - (n - y);
  if (u > 0) ans -= fun(u);
  if (d > 0) ans -= fun(d);
  if (l > 0) ans -= fun(l);
  if (r > 0) ans -= fun(r);
  if (u > 0 && l > 0) ans += sum(u, l, t + 1);
  if (d > 0 && l > 0) ans += sum(d, l, t + 1);
  if (u > 0 && r > 0) ans += sum(u, r, t + 1);
  if (d > 0 && r > 0) ans += sum(d, r, t + 1);
  return ans;
}
long long int bsearc(long long int n, long long int x, long long int y,
                     long long int c) {
  long long int l = 0, h = 2 * n + 1, m;
  while (l < h) {
    m = (l + h) / 2;
    if (cal(n, x, y, m) < c)
      l = m + 1;
    else
      h = m;
  }
  return l;
}
int main() {
  long long int n, x, y, c;
  cin >> n >> x >> y >> c;
  cout << bsearc(n, x, y, c) << "\n";
  return 0;
}
