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, c;
long long rx, ry, cx, cy;
int trans(long long xx) { return xx * (xx + 1) / 2; }
bool check(long long w) {
  long long hig = min(w, rx) + min(w, ry) + 1,
            wid = min(w, cx) + min(w, cy) + 1;
  long long sum = hig * wid;
  long long u_l, u_r, d_l, d_r;
  if (w <= rx) u_l = max(0ll, min(w, cx));
  if (rx < w) u_l = max(0ll, min(w, cx) - (w - rx));
  if (w <= rx) u_r = max(0ll, min(w, cy));
  if (rx < w) u_r = max(0ll, min(w, cy) - (w - rx));
  if (w <= ry) d_l = max(0ll, min(w, cx));
  if (ry < w) d_l = max(0ll, min(w, cx) - (w - ry));
  if (w <= ry) d_r = max(0ll, min(w, cy));
  if (ry < w) d_r = max(0ll, min(w, cy) - (w - ry));
  return sum - trans(u_l) - trans(u_r) - trans(d_l) - trans(d_r) >= c;
}
int main() {
  cin >> n >> x >> y >> c;
  rx = x - 1, ry = n - x, cx = y - 1, cy = n - y;
  long long l = 0, r = 1e9;
  while (l < r) {
    long long mid = (l + r) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid + 1;
  }
  cout << r;
}
