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 int n, x, y, c;
long long int core(long long int t) {
  long long int fin = 1;
  for (int i = 0; i < 4; i++) {
    if (x + y < t)
      fin += x * (y + 1);
    else {
      fin += t * (t + 1) / 2;
      if (x < t) fin -= (t - x) * (t - x + 1) / 2;
      if (y < t) fin -= (t - y - 1) * (t - y) / 2;
    }
    swap(x, y);
    x = n - 1 - x;
  }
  return fin;
}
int main() {
  scanf("%d %d %d %d", &n, &x, &y, &c);
  y--;
  x--;
  long long int low = -1;
  long long int high = 2 * n;
  long long int mid;
  while (low + 1 < high) {
    mid = (low + high) / 2;
    if (core(mid) >= c)
      high = mid;
    else
      low = mid;
  }
  cout << high;
  return 0;
}
