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;
const int N = 1e5 + 5, Mod = 1e9 + 7;
long long int pw1(long long int x) {
  if (x > 0) return x * x;
  return 0LL;
}
long long int pw2(long long int x) {
  if (x > 0) return x * (x + 1) / 2;
  return 0LL;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, x, y, c;
  cin >> n >> x >> y >> c;
  long long int lo = 0, hi = 2 * n;
  --x, --y, --n;
  while (lo < hi) {
    long long int mid = (lo + hi) >> 1;
    long long int Now = 2 * mid * (mid + 1) + 1;
    Now -= pw1(mid - x);
    Now -= pw1(mid - y);
    Now -= pw1(mid + x - n);
    Now -= pw1(mid + y - n);
    Now += pw2(mid - x - y - 1);
    Now += pw2(mid - n + x - y - 1);
    Now += pw2(mid - n - x + y - 1);
    Now += pw2(mid - 2 * n + x + y - 1);
    if (Now >= c)
      hi = mid;
    else
      lo = mid + 1;
  }
  cout << lo << "\n";
  return 0;
}
