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() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, x, y, c;
  cin >> n >> x >> y >> c;
  long long l = 0, r = 2 * n + 100;
  long long left = y, right = n + 1 - y, up = x, down = n + 1 - x;
  if (c == 1) {
    cout << 0 << '\n';
    return 0;
  }
  while (l + 1 < r) {
    long long m = (l + r) >> 1ll;
    long long f = 1 + 2 * m * (m + 1);
    if (left <= m) f -= (m - left + 1) * (m - left + 1);
    if (right <= m) f -= (m - right + 1) * (m - right + 1);
    if (up <= m) f -= (m - up + 1) * (m - up + 1);
    if (down <= m) f -= (m - down + 1) * (m - down + 1);
    if (right + up <= m)
      f += (m - (right + up) + 2) * (m - (right + up) + 1) / 2;
    if (left + up <= m) f += (m - (left + up) + 2) * (m - (left + up) + 1) / 2;
    if (right + down <= m)
      f += (m - (right + down) + 2) * (m - (right + down) + 1) / 2;
    if (left + down <= m)
      f += (m - (left + down) + 2) * (m - (left + down) + 1) / 2;
    if (f < c)
      l = m;
    else
      r = m;
  }
  cout << r << '\n';
  return 0;
}
