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 sum, mid, n, x, y, c, l, r, ans;
long long go(long long mid) {
  sum = 0;
  sum += mid * mid + (mid - 1) * (mid - 1);
  if (mid > x) sum -= (mid - x) * (mid - x);
  if (mid > y) sum -= (mid - y) * (mid - y);
  if (mid > (n - x) + 1) sum -= (mid - (n - x) - 1) * (mid - (n - x) - 1);
  if (mid > (n - y) + 1) sum -= (mid - (n - y) - 1) * (mid - (n - y) - 1);
  if (mid > x + y) sum += (mid - x - y) * (mid - x - y + 1) / 2;
  if (mid > y + n - x + 1)
    sum += (mid - y - (n - x) - 1) * (mid - y - (n - x)) / 2;
  if (mid > x + (n - y) + 1)
    sum += (mid - x - (n - y) - 1) * (mid - x - (n - y)) / 2;
  if (mid > (n - x) + (n - y) + 2)
    sum += (mid - (n - x) - (n - y) - 2) * (mid - (n - x) - (n - y) - 1) / 2;
  if (sum >= c) return 1;
  return 0;
}
int main() {
  cin >> n >> x >> y >> c;
  if (c == 1) {
    cout << 0;
    return 0;
  }
  l = 0;
  r = 2000000000;
  while (l <= r) {
    mid = (l + r) / 2;
    if (go(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << ans - 1;
}
