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;
int main() {
  int n, x, y, c, d1, d2, d3, d4, l, r, u, d, i = 0, c1, c2, c3, c4;
  cin >> n >> x >> y >> c;
  d1 = min(n - y, x - 1);
  c1 = max(n - y, x - 1) - d1;
  d2 = min(n - y, n - x);
  c2 = max(n - y, n - x) - d2;
  d3 = min(y - 1, x - 1);
  c3 = max(y - 1, x - 1) - d3;
  d4 = min(y - 1, n - x);
  c4 = max(y - 1, n - x) - d4;
  long sum = 1;
  u = x, d = x, l = y, r = y;
  while (sum < c) {
    if (i) {
      if (i > d1 && i <= c1 + d1)
        sum += d1;
      else if (i <= d1)
        sum += i;
      else
        sum += max(0, --d1);
      if (i > d2 && i <= c2 + d2)
        sum += d2;
      else if (i <= d2)
        sum += i;
      else
        sum += max(0, --d2);
      if (i > d3 && i <= c3 + d3)
        sum += d3;
      else if (i <= d3)
        sum += i;
      else
        sum += max(0, --d3);
      if (i > d4 && i <= c4 + d4)
        sum += d4;
      else if (i <= d4)
        sum += i;
      else
        sum += max(0, --d4);
    }
    if (u > 1) sum++;
    if (d < n) sum++;
    if (l > 1) sum++;
    if (r < n) sum++;
    u--, d++, l--, r++;
    i++;
  }
  cout << i;
  return 0;
}
