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>
const long double PI = 3.141592653589793238L;
const long long N = 300005;
using namespace std;
long long sum(long long a, long long b) {
  return (b >= a) ? (((b - a + 1) * (b + a)) / 2) : 0ll;
}
long long f(long long a, long long x, long long b, long long mn) {
  if (mn >= x) return sum(a, x) + sum(b, x) - x;
  if (mn >= max(a, b))
    return ((2 * (x - mn + 1) - 1) * mn) + sum(a, mn - 1) + sum(b, mn - 1);
  if (mn <= min(a, b)) return ((x - a + 1) + (x - b + 1) - 1) * mn;
  if (mn < b) return sum(a, mn) + f(mn + 1, x, b, mn);
  if (mn < a) return sum(b, mn) + f(a, x, mn + 1, mn);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, x, y, c;
  cin >> n >> x >> y >> c;
  long long i = 0, j = 2e18;
  long long ans;
  while (i <= j) {
    long long mid = (i + j) / 2;
    long long left = min(y - 1, mid);
    long long right = min(n - y, mid);
    long long top = f(mid - left, mid, mid - right, x - 1);
    long long down = f(mid - left, mid, mid - right, n - x);
    if (top + down + left + right + 1 >= c)
      ans = mid, j = mid - 1;
    else
      i = mid + 1;
  }
  cout << ans << '\n';
  return 0;
}
