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;
const long long M = 2e5 + 5;
const long long INF = 1e15 + 5;
const long long MOD = 1e9 + 7;
const long long N = 1e5 + 5;
const double PI = 3.14159265358979323846264338327950288419716939937510;
long long n, x, y, c;
long long gi(long long g) {
  if (g <= 0) return 0;
  return g * g;
}
long long gj(long long g) {
  if (g <= 0) return 0;
  return (g * g - g) / 2 + g;
}
long long gett(long long t) {
  long long ss = (2 * t + 1) * (t * 2 + 1);
  ss = ss / 2 + 1;
  ss -= gi(t - (x - 1));
  ss -= gi(t - (n - x));
  ss -= gi(t - (y - 1));
  ss -= gi(t - (n - y));
  ss += gj(t - (x - 1) - (y - 1) - 1);
  ss += gj(t - (n - x) - (n - y) - 1);
  ss += gj(t - (n - x) - (y - 1) - 1);
  ss += gj(t - (x - 1) - (n - y) - 1);
  return ss;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> x >> y >> c;
  long long low = 0, high = 2e9;
  while (low < high) {
    long long mid = low + (high - low) / 2;
    if (gett(mid) >= c)
      high = mid;
    else
      low = mid + 1;
  }
  cout << low << "\n";
}
