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 howMany(long long x, long long y, long long turn, long long size) {
  long long result = (2 + 2 * turn) * turn;
  if (x + turn > size) {
    result -= (x + turn - size) * (x + turn - size);
  }
  if (x - turn < 1) {
    long long temp = abs(x - turn - 1);
    result -= temp * temp;
  }
  if (y + turn > size) {
    result -= (y + turn - size) * (y + turn - size);
    long long temp = y + turn - size;
    if (x + temp - 1 > size) {
      long long temp2 = x + temp - 1 - size;
      result += ((temp2 + 1) * temp2) / 2;
    }
    if (x - temp + 1 < 1) {
      long long temp2 = abs(x - temp);
      result += ((temp2 + 1) * temp2) / 2;
    }
  }
  if (y - turn < 1) {
    long long temp = abs(y - turn - 1);
    result -= temp * temp;
    if (x + temp - 1 > size) {
      long long temp2 = x + temp - 1 - size;
      result += ((temp2 + 1) * temp2) / 2;
    }
    if (x - temp + 1 < 1) {
      long long temp2 = abs(x - temp);
      result += ((temp2 + 1) * temp2) / 2;
    }
  }
  return result;
}
int main() {
  long long n, x, y, c;
  cin >> n >> x >> y >> c;
  --c;
  long long left = 0, right = 2 * n;
  long long answer;
  while (left < right) {
    long long middle = (right + left) / 2;
    if (howMany(x, y, middle, n) >= c) {
      answer = middle;
      right = middle;
    } else {
      left = middle + 1;
    }
  }
  cout << answer;
  return 0;
}
