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 n, x, y, c;
  while (cin >> n >> x >> y >> c) {
    long long r[4] = {y - 1, x - 1, n - y, n - x};
    long long t = 0;
    long long cnt = 1;
    for (;;) {
      for (int i = 0; i < 4; i++) {
        cnt +=
            t - min(max(0LL, t - r[i]) + max(0LL, t - 1 - r[(i + 1) % 4]), t);
      }
      if (cnt >= c) break;
      ++t;
    }
    cout << t << endl;
  }
}
