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, u, d;
long long n, x, y, c;
long long sqr(long long x) { return x * x; }
long long tri(long long x) { return (x + 1) * x / 2; }
bool solve(long long t) {
  long long sum = t * t + (t + 1) * (t + 1);
  if (t > l) sum -= sqr(t - l);
  if (t > r) sum -= sqr(t - r);
  if (t > u) sum -= sqr(t - u);
  if (t > d) sum -= sqr(t - d);
  if (t > l + d) sum += tri(t - (l + d) - 1);
  if (t > l + u) sum += tri(t - (l + u) - 1);
  if (t > r + d) sum += tri(t - (r + d) - 1);
  if (t > r + u) sum += tri(t - (r + u) - 1);
  if (sum >= c) return true;
  return false;
}
int main() {
  while (~scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c)) {
    l = x - 1, r = n - x;
    u = y - 1, d = n - y;
    long long low = 0, high = 2 * n, mid;
    while (low <= high) {
      mid = (low + high) / 2;
      if (solve(mid))
        high = mid - 1;
      else
        low = mid + 1;
    }
    printf("%I64d\n", low);
  }
  return 0;
}
