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 func2(long long x, long long y, long long t) {
  if (t >= x + y - 2) return x * y;
  long long ans = (t + 1) * (t + 2) / 2;
  if (t >= x) ans -= (t - x + 1) * (t - x + 2) / 2;
  if (t >= y) ans -= (t - y + 1) * (t - y + 2) / 2;
  return ans;
}
long long func(long long N, long long x, long long y, long long t) {
  long long ans = func2(x, y, t) + func2(x, N - y + 1, t) +
                  func2(N - x + 1, y, t) + func2(N - x + 1, N - y + 1, t);
  ans -= 3;
  ans -= min(x - 1, t);
  ans -= min(N - x, t);
  ans -= min(y - 1, t);
  ans -= min(N - y, t);
  return ans;
}
int main(void) {
  long long N, x, y, c;
  cin >> N >> x >> y >> c;
  if (1 >= c) {
    cout << 0 << endl;
    return 0;
  }
  long long low = 0, high = (1ll << 33);
  while (high - low > 1) {
    long long mid = (high + low) / 2;
    long long tmp = func(N, x, y, mid);
    if (tmp >= c)
      high = mid;
    else
      low = mid;
  }
  cout << high << endl;
  return 0;
}
