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 ans;
long long x, y, n, t;
long long geit(long long u) {
  long long total = 2 * (u + 1) * u + 1;
  if (u > x - 1) {
    total -= (u + 1 - x) * (u + 1 - x);
    if (u > n + x - y) total += (u - n - x + y + 1) * (u - n - x + y) / 2;
  }
  if (u > y - 1) {
    total -= (u + 1 - y) * (u + 1 - y);
    if (u > x + y - 1) total += (u - x - y + 2) * (u - x - y + 1) / 2;
  }
  if (u > n - x) {
    total -= (u + x - n) * (u + x - n);
    if (u > n - x + y) total += (u - n + x - y) * (u - n + x - y + 1) / 2;
  }
  if (u > n - y) {
    total -= (u + y - n) * (u + y - n);
    if (u > 2 * n - x - y + 1)
      total += (u - 2 * n + x + y) * (u - 2 * n + x + y - 1) / 2;
  }
  return total;
}
int main() {
  while (cin >> n >> x >> y >> t) {
    long long mid = 0;
    while (geit(mid) < t) mid++;
    cout << mid << endl;
  }
  return 0;
}
