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;
int n, x, y, c, l, r;
long long kol, mid;
long long P(long long n) { return n * n; }
long long D(long long n) {
  if (n < 0)
    return 0;
  else
    return n * (n + 1) / 2;
}
int main(void) {
  scanf("%d%d%d%d", &n, &x, &y, &c);
  if (c <= 1) {
    printf("0");
    return 0;
  }
  l = 0;
  r = n + n;
  for (int iter = 0; iter < 100; iter++) {
    mid = (l + r) / 2;
    kol = (mid + mid) * (mid + 1) + 1;
    if (mid > x - 1) kol -= P(mid - x + 1);
    if (mid > y - 1) kol -= P(mid - y + 1);
    if (mid > n - x) kol -= P(mid - n + x);
    if (mid > n - y) kol -= P(mid - n + y);
    if (mid > x - 1 && mid > y - 1) kol += D(mid - y + 1 - x);
    if (mid > x - 1 && mid > n - y) kol += D(mid - n + y - x);
    if (mid > n - x && mid > y - 1) kol += D(mid - y + 1 - (n - x + 1));
    if (mid > n - x && mid > n - y) kol += D(mid - n + y - (n - x + 1));
    if (kol >= c)
      r = mid - 1;
    else
      l = mid;
  }
  l += 100;
  kol = c;
  while (kol >= c) {
    l--;
    mid = l;
    kol = (mid + mid) * (mid + 1) + 1;
    if (mid > x - 1) kol -= P(mid - x + 1);
    if (mid > y - 1) kol -= P(mid - y + 1);
    if (mid > n - x) kol -= P(mid - n + x);
    if (mid > n - y) kol -= P(mid - n + y);
    if (mid > x - 1 && mid > y - 1) kol += D(mid - y + 1 - x);
    if (mid > x - 1 && mid > n - y) kol += D(mid - n + y - x);
    if (mid > n - x && mid > y - 1) kol += D(mid - y + 1 - (n - x + 1));
    if (mid > n - x && mid > n - y) kol += D(mid - n + y - (n - x + 1));
  }
  printf("%d\n", l + 1);
  return 0;
}
