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;
long long getU(long long t, long long i, long long j) {
  if (t <= i) return 0;
  return (t - i) * (t - i);
}
long long getL(long long t, long long i, long long j) {
  if (t <= j) return 0;
  long long ret = (t - j) * (t - j);
  long long mni = i + j - (-1) - t;
  mni *= -1;
  if (mni > 0) ret -= mni * (mni + 1) / 2;
  long long mxi = t + (-1) - j + i;
  mxi -= n - 1;
  if (mxi > 0) ret -= mxi * (mxi + 1) / 2;
  return ret;
}
bool ok(long long t) {
  return 1 + 2 * t * (t + 1) - getU(t, x, y) - getU(t, n - 1 - x, y) -
             getL(t, x, y) - getL(t, x, n - 1 - y) >=
         c;
}
int main() {
  cin >> n >> x >> y >> c;
  --x, --y;
  long long s = 0, e = 2 * n, md, bst;
  while (s <= e) {
    md = s + ((e - s) >> 1);
    if (ok(md))
      e = md - 1, bst = md;
    else
      s = md + 1;
  }
  cout << bst;
}
