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, C, X, Y;
long long cal(long long int t) {
  long long a[4] = {min(X - 1, t), min(N - Y, t), min(N - X, t), min(Y - 1, t)};
  long long int r;
  r = 2 * (t + 1) * (t + 1) - (2 * t + 1);
  for (int i = 0; i < 4; i++) {
    r -= (t - a[i]) * (t - a[i]);
    if ((a[i] + a[(i + 1) % 4]) >= t) continue;
    r += (t - a[i] - a[(i + 1) % 4] - 1) * (t - a[i] - a[(i + 1) % 4]) / 2;
  }
  return r;
}
int main() {
  cin >> N >> X >> Y >> C;
  if (cal(0) >= C) {
    cout << 0 << endl;
    return 0;
  }
  int l = 0, r = N * 2, c = (l + r) / 2;
  while (l != c) {
    long long int t = cal(c);
    if (t >= C)
      r = c;
    else
      l = c;
    c = (l + r) / 2;
  }
  if (cal(c) >= C)
    cout << c << endl;
  else
    cout << r << endl;
  return 0;
}
