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 x, y, n, c;
bool check(long long t) {
  long long ans = 2 * t * t + 2 * t + 1;
  if (x + t > n) {
    long long m = x + t - n;
    ans -= m * m;
  }
  if (y + t > n) {
    long long m = y + t - n;
    ans -= m * m;
  }
  if (x - 1 < t) {
    long long m = t - x + 1;
    ans -= m * m;
  }
  if (y - 1 < t) {
    long long m = t - y + 1;
    ans -= m * m;
  }
  if (x - 1 + y - 1 < t) {
    long long a = t - (x - 1 + y - 1) - 1;
    ans += a * (a + 1) / 2;
  }
  if (x - 1 + n - y < t) {
    long long a = t - (x - 1 + n - y) - 1;
    ans += a * (a + 1) / 2;
  }
  if (n - x + y - 1 < t) {
    long long a = t - (n - x + y - 1) - 1;
    ans += a * (a + 1) / 2;
  }
  if (n - x + n - y < t) {
    long long a = t - (n - x + n - y) - 1;
    ans += a * (a + 1) / 2;
  }
  return ans >= c;
}
int main() {
  while (cin >> n >> x >> y >> c) {
    long long low = 0, high = 2 * n, mid, ans = 0;
    while (low < high) {
      mid = (low + high) / 2;
      if (check(mid)) {
        ans = mid;
        high = mid;
      } else
        low = mid + 1;
    }
    cout << ans << endl;
  }
  return 0;
}
