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;
bool check(long long mid) {
  long long x1 = max(1ll, x - mid);
  long long x2 = min(n, x + mid);
  long long y1 = min(n, y + mid);
  long long y2 = max(1ll, y - mid);
  long long ans = (x2 - x1 + 1ll) * (y1 - y2 + 1ll);
  long long a = x - mid;
  long long b = y + max(0ll, x1 - a) + 1ll;
  if (b <= y1) ans -= ((y1 - b + 1ll) * (y1 - b + 2ll) / 2ll);
  a = x - mid;
  b = y - max(0ll, x1 - a) - 1ll;
  if (b >= y2) ans -= ((b - y2 + 1ll) * (b - y2 + 2ll) / 2ll);
  a = x + mid;
  b = y + max(0ll, a - x2) + 1ll;
  if (b <= y1) ans -= ((y1 - b + 1ll) * (y1 - b + 2ll) / 2ll);
  a = x + mid;
  b = y - max(0ll, a - x2) - 1ll;
  if (b >= y2) ans -= ((b - y2 + 1ll) * (b - y2 + 2ll) / 2ll);
  return (ans >= c);
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0, r = c;
  long long ans = r;
  while (l <= r) {
    long long mid = (l + r) / 2ll;
    if (check(mid))
      r = mid - 1, ans = min(ans, mid);
    else
      l = mid + 1;
  }
  return cout << ans << "\n", 0;
}
