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 n, x, y, ans = -1;
long long c;
long long maxx, maxy, minx, miny;
long long s;
int main() {
  cin >> n >> x >> y >> c;
  while (s < c) {
    ans++;
    maxx = x + ans;
    maxy = y + ans;
    minx = x - ans;
    miny = y - ans;
    s = 2 * ans * ans + (ans * 2 + 1);
    long long t1, t2, t3, t4;
    t1 = max(maxx - n, 0LL);
    t2 = max(maxy - n, 0LL);
    t3 = max(1 - minx, 0LL);
    t4 = max(1 - miny, 0LL);
    s -= t1 * t1;
    s -= t2 * t2;
    s -= t3 * t3;
    s -= t4 * t4;
    long long tt1, tt2, tt3, tt4;
    tt1 = max(t1 - (n + 1 - y), 0LL);
    tt2 = max(t1 - y, 0LL);
    tt3 = max(t3 - y, 0LL);
    tt4 = max(t3 - (n + 1 - y), 0LL);
    s += tt1 * (tt1 + 1LL) / 2LL;
    s += tt2 * (tt2 + 1LL) / 2LL;
    s += tt3 * (tt3 + 1LL) / 2LL;
    s += tt4 * (tt4 + 1LL) / 2LL;
  }
  cout << ans;
  return 0;
}
