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>
long long X, Y, N, C;
long long l, r, m, o, v;
long long triang(long long x) {
  if (x <= 0) return 0;
  return x * (x + 1) / 2;
}
long long skvr(long long x) {
  if (x <= 0) return 0;
  return x * x;
}
long long cornerloss(long long x, long long y, long long d) {
  return triang(d - x - y + 1);
}
long long sideloss(long long x, long long d) { return skvr(d - x + 1); }
long long noloss(long long d) { return d * d + (1 + d) * (1 + d); }
long long totsrcf(long long x, long long y, long long d) {
  return noloss(d) - sideloss(x, d) - sideloss(y, d) - sideloss(N - x + 1, d) -
         sideloss(N - y + 1, d) + cornerloss(x, y, d) +
         cornerloss(N - x + 1, y, d) + cornerloss(N - x + 1, N - y + 1, d) +
         cornerloss(x, N - y + 1, d);
}
void debug() {
  scanf("%I64d", &m);
  printf("%I64d\n", totsrcf(X, Y, m));
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &N, &X, &Y, &C);
  l = 0;
  r = 2000000000ll;
  o = 0;
  while (l <= r) {
    m = (l + r) / 2;
    v = totsrcf(X, Y, m);
    if (v >= C) {
      o = m;
      r = m - 1;
    } else {
      l = m + 1;
    }
  }
  printf("%I64d\n", o);
  return 0;
}
