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;
int u, d, l, r;
int ur, rd, dl, lu;
int cnt, step;
int main() {
  cin >> n >> x >> y >> c;
  cnt = 1;
  u = d = x;
  l = r = y;
  while (cnt < c) {
    step++;
    if (u != 0) u--;
    if (d != n + 1) d++;
    if (l != 0) l--;
    if (r != n + 1) r++;
    if (u != 0) cnt++;
    if (d != n + 1) cnt++;
    if (l != 0) cnt++;
    if (r != n + 1) cnt++;
    cnt += ur;
    cnt += rd;
    cnt += dl;
    cnt += lu;
    ur++;
    rd++;
    dl++;
    lu++;
    if (u == 0) lu--;
    if (u == 0) ur--;
    if (d == n + 1) rd--;
    if (d == n + 1) dl--;
    if (l == 0) dl--;
    if (l == 0) lu--;
    if (r == n + 1) ur--;
    if (r == n + 1) rd--;
    if (ur < 0) ur = 0;
    if (rd < 0) rd = 0;
    if (dl < 0) dl = 0;
    if (lu < 0) lu = 0;
  }
  cout << step << '\n';
  return 0;
}
