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;
long long c;
long long compute(long long x, long long y, long long mid) {
  long long L = mid;
  if ((n - x) + (n - y) <= L) return (((long long)n - x + 1) * (n - y + 1));
  long long ret = (L + 2) * (L + 1) / 2;
  if (x + mid >= n)
    ret -= ((long long)x + mid - n) * ((long long)x + mid - n + 1) / 2;
  if (y + mid >= n)
    ret -= ((long long)y + mid - n) * ((long long)y + mid - n + 1) / 2;
  return ret;
}
long long check(long long mid) {
  if (mid == 0) return 1;
  long long ret = compute(x, y, mid) + compute(n - x + 1, y, mid) +
                  compute(x, n - y + 1, mid) +
                  compute(n + 1 - x, n + 1 - y, mid);
  ret -= min(n, x + mid) - max(1ll, x - mid) + 1;
  ret -= min(n, y + mid) - max(1ll, y - mid) + 1;
  ret -= 1;
  return ret;
}
int main() {
  cin >> n >> x >> y >> c;
  long long g = 0, h = 2 * n, ans = 0;
  while (g <= h) {
    int mid = (g + h) / 2;
    if (check(mid) >= c)
      ans = mid, h = mid - 1;
    else
      g = mid + 1;
  }
  cout << ans << endl;
  return 0;
}
