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, x, y, c;
long long colored(long long n) { return n * n + (n - 1) * (n - 1); }
bool valid(long long t) {
  long long colored = 1LL * t * t + (t - 1LL) * (t - 1LL);
  long long right = 0, left = 0, top = 0, bottom = 0;
  if (y + t > n) right = y + t - n - 1;
  if (y - t < 1) left = 1 - y + t - 1;
  if (x + t > n) bottom = x + t - n - 1;
  if (x - t < 1) top = 1 - x + t - 1;
  colored -= right * right + left * left + top * top + bottom * bottom;
  long long tr = right - x;
  long long tl = left - x;
  long long br = right - (n - x + 1);
  long long lr = left - (n - x + 1);
  if (tr > 0) colored += tr * (tr + 1) / 2;
  if (tl > 0) colored += tl * (tl + 1) / 2;
  if (br > 0) colored += br * (br + 1) / 2;
  if (lr > 0) colored += lr * (lr + 1) / 2;
  return colored >= c;
}
int main() {
  cin >> n >> x >> y >> c;
  long long lo = 1, hi = n * 2;
  while (lo < hi) {
    int mid = (lo + hi) / 2;
    if (valid(mid))
      hi = mid;
    else
      lo = mid + 1;
  }
  cout << lo - 1 << endl;
  return 0;
}
