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 s1(long long n) {
  if (n <= 0) return 0;
  return n * n;
}
long long s2(long long n) {
  if (n <= 0) return 0;
  return n * (n + 1) / 2;
}
long long check(long long m) {
  long long ret = 0;
  ret += s1(m) + s1(m + 1);
  ret -= s1(m + x - n) + s1(m + y - n) + s1(m - x + 1) + s1(m - y + 1);
  ret += s2(m - (n - x + n - y) - 1);
  ret += s2(m - (n - x + y - 1) - 1);
  ret += s2(m - (x - 1 + n - y) - 1);
  ret += s2(m - (x - 1 + y - 1) - 1);
  return ret;
}
long long binary() {
  long long s = 0, e = n * 2;
  while (s != e) {
    long long m = (s + e) / 2;
    if (check(m) >= c)
      e = m;
    else
      s = m + 1;
  }
  return e;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> x >> y >> c;
  cout << binary() << endl;
  return 0;
}
