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 licz1(long long x) {
  if (x < 0) return 0;
  return x * (x - 1) + x;
}
long long licz2(long long x) {
  if (x < 0) return 0;
  return (1 + x) * x / 2;
}
long long Ile(long long n, long long x, long long y, long long t) {
  long long il = 2 * t * t + 2 * t + 1;
  il -= licz1(t + 1 - x);
  il -= licz1(t + 1 - y);
  il -= licz1(t + 1 - (n - x + 1));
  il -= licz1(t + 1 - (n - y + 1));
  il += licz2(t + 1 - x - (n - y + 1));
  il += licz2(t + 1 - x - y);
  il += licz2(t + 1 - (n - x + 1) - (n - y + 1));
  il += licz2(t + 1 - (n - x + 1) - y);
  return il;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long n, x, y, c;
  cin >> n >> x >> y >> c;
  long long l = 0, p = 4 * n, s, r = -1;
  while (l <= p) {
    s = (l + p) / 2;
    if (Ile(n, x, y, s) >= c) {
      r = s;
      p = s - 1;
    } else
      l = s + 1;
  }
  cout << r;
  return 0;
}
