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 l, r, n, x, y, c, d[5];
long long pravi(long long t) { return 1 + (t * (t + 1) / 2) * 4; }
long long skida(long long x) { return x > 0 ? 1 + --x + x * (x + 1) : 0; }
long long preklop(long long p) { return p > 0 ? p * (p + 1) / 2 : 0; }
int main() {
  cin >> n >> x >> y >> c;
  for (l = 0, r = 1e9; l < r;) {
    long long mid = (l + r) / 2;
    long long temp = pravi(mid);
    d[0] = max(0LL, x + mid - n);
    d[1] = max(0LL, y + mid - n);
    d[2] = max(0LL, mid + 1 - x);
    d[3] = max(0LL, mid + 1 - y);
    for (int i = 0; i < 4; i++) temp -= skida(d[i]);
    for (int i = 0; i < 4; i++)
      temp += preklop(d[i] + d[(i + 1) & 3] - mid - 1);
    if (temp < c)
      l = mid + 1;
    else
      r = mid;
  }
  cout << l;
  return 0;
}
