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 l, r, u, d, n, x, y, c;
bool judge(long long t) {
  long long ans = t * t + (t + 1) * (t + 1);
  if (t > l) ans -= (t - l) * (t - l);
  if (t > r) ans -= (t - r) * (t - r);
  if (t > u) ans -= (t - u) * (t - u);
  if (t > d) ans -= (t - d) * (t - d);
  if (t > l + d) ans += (t - l - d - 1) * (t - l - d) / 2;
  if (t > l + u) ans += (t - l - u - 1) * (t - l - u) / 2;
  if (t > r + d) ans += (t - r - d - 1) * (t - r - d) / 2;
  if (t > r + u) ans += (t - r - u - 1) * (t - r - u) / 2;
  if (ans >= c) return 1;
  return 0;
}
int main() {
  cin >> n >> x >> y >> c;
  l = x - 1;
  r = n - x;
  u = y - 1;
  d = n - y;
  long long le = 0, ri = 2 * n, mid;
  while (le <= ri) {
    mid = (le + ri) / 2;
    if (judge(mid))
      ri = mid - 1;
    else
      le = mid + 1;
  }
  cout << le << endl;
  return 0;
}
