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() {
  long long x, y, n, c;
  scanf("%I64d %I64d %I64d %I64d", &n, &x, &y, &c);
  long long hi = c / 2, low = 0, mid = (hi + low) / 2;
  int l = 0, r = 0, t = 0, b = 0;
  float s = 1.0;
  while (mid < hi) {
    s = mid * mid * 2 + mid * 2 + 1;
    l = x + mid - n, r = y + mid - n, t = mid - x + 1, b = mid - y + 1;
    if (x + mid > n) {
      s -= l * l;
    }
    if (y + mid > n) {
      s -= r * r;
    }
    if (mid - x > -1) {
      s -= t * t;
    }
    if (mid - y > -1) {
      s -= b * b;
    }
    l = mid - x - y + 1, r = mid - n + x - y, t = mid - n + y - x,
    b = mid - 2 * n + x + y - 1;
    if (l > 0) {
      s += 0.5 * l * l + 0.5 * l;
    }
    if (r > 0) {
      s += 0.5 * r * r + 0.5 * r;
    }
    if (t > 0) {
      s += 0.5 * t * t + 0.5 * t;
    }
    if (b > 0) {
      s += 0.5 * b * b + 0.5 * b;
    }
    if (s >= c)
      hi = mid;
    else
      low = mid + 1;
    mid = (hi + low) / 2;
  }
  printf("%I64d", mid);
}
