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, flag, tx, ty, a, ans;
bool check(long long t) {
  long long sum = t * (t + 1) * 2 + 1, x, y, z;
  x = t - tx + 1;
  if (x > 0) sum -= x * (x + 1) / 2 + (x - 1) * x / 2;
  x = t - ty + 1;
  if (x > 0) sum -= x * (x + 1) / 2 + (x - 1) * x / 2;
  x = t + tx - n;
  if (x > 0) sum -= x * (x + 1) / 2 + (x - 1) * x / 2;
  x = t + ty - n;
  if (x > 0) sum -= x * (x + 1) / 2 + (x - 1) * x / 2;
  x = t - tx + 1, y = n - ty + 1, z = x - y;
  if (z > 0) sum += z * (z + 1) / 2;
  x = t - ty + 1, y = tx, z = x - y;
  if (z > 0) sum += z * (z + 1) / 2;
  x = t + tx - n, y = ty, z = x - y;
  if (z > 0) sum += z * (z + 1) / 2;
  x = t + ty - n, y = n - tx + 1, z = x - y;
  if (z > 0) sum += z * (z + 1) / 2;
  if (sum >= a) return 1;
  return 0;
}
signed main() {
  cin >> n >> tx >> ty >> a;
  long long l = 0, r = n * 2 + 1;
  while (l <= r) {
    long long mid = (l + r) / 2;
    if (check(mid))
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  cout << ans;
}
