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 n, row, col, c;
long long over_top_bot(long long over) { return over * over; }
long long over_left_right(long long over) {
  long long base = over * over;
  long long top_bad = (over >= row) ? ((over - row) * (over - row + 1)) / 2 : 0;
  long long bot_bad =
      (over >= n - row + 1)
          ? ((over - (n - row + 1)) * (over - (n - row + 1) + 1)) / 2
          : 0;
  return base - top_bad - bot_bad;
}
long long total_area(long long h) {
  long long quarter = (h * (h + 1)) / 2;
  return 4LL * quarter - 4LL * h + 1;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin >> n >> row >> col >> c;
  long long bot = 0;
  long long top = 2LL * n;
  while (bot + 1 != top) {
    long long mid = (bot + top) / 2LL;
    long long over_top = max(0LL, mid - row);
    long long over_bot = max(0LL, mid - (n - row + 1));
    long long over_left = max(0LL, mid - col);
    long long over_right = max(0LL, mid - (n - col + 1));
    long long all = total_area(mid) - over_top_bot(over_top) -
                    over_top_bot(over_bot) - over_left_right(over_left) -
                    over_left_right(over_right);
    if (all >= c)
      top = mid;
    else
      bot = mid;
  }
  cout << top - 1 << endl;
  return 0;
}
