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 ct(long long a, long long b, long long t) {
  b = min(b, t + 1);
  a = min(a, t + 1);
  if (b <= t + 2 - a) {
    return a * b;
  }
  long long ret = (t + 2 - a) * a;
  b -= (t + 2 - a);
  b = min(a - 1, b);
  ret += (a - 1 + a - b) * (b) / 2;
  return ret;
}
bool check(long long t) {
  long long ret = 0;
  ret += ct(x, n + 1 - y, t);
  ret += ct(n + 1 - x, n + 1 - y, t);
  ret += ct(y, x, t);
  ret += ct(y, n + 1 - x, t);
  ret -= min(t + 1, x) + min(t + 1, n + 1 - x);
  ret -= min(t + 1, y) + min(t + 1, n + 1 - y);
  ret += 1;
  return ret >= c;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> x >> y >> c;
  check(1);
  long long l = 0, r = n * n;
  while (l < r) {
    long long m = (l + r) / 2;
    if (check(m)) {
      r = m;
    } else {
      l = m + 1;
    }
  }
  cout << l << endl;
  return 0;
}
