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 main(int argc, char const *argv[]) {
  int n, x, y, c;
  std::cin >> n >> x >> y >> c;
  ++n;
  bool topLeft = true, topRight = true, bottomLeft = true, bottomRight = true;
  int counter = 1, timer = 0, arm = 1, up = 0, left = 0, down = 0, right = 0;
  int xiterator, yiterator;
  while (counter < c) {
    if (topLeft) {
      xiterator = x - up;
      yiterator = y + arm;
      if (yiterator >= n) {
        yiterator = n - 1;
        ++up;
        --xiterator;
      }
      while (xiterator > 0 && yiterator > y) {
        if (xiterator == 1 && yiterator == n - 1) topLeft = false;
        ++counter;
        --xiterator;
        --yiterator;
      }
    }
    if (bottomLeft) {
      xiterator = x - arm;
      yiterator = y - left;
      if (xiterator <= 0) {
        xiterator = 1;
        ++left;
        --yiterator;
      }
      while (xiterator < x && yiterator > 0) {
        if (xiterator == 1 && yiterator == 1) bottomLeft = false;
        ++counter;
        ++xiterator;
        --yiterator;
      }
    }
    if (bottomRight) {
      xiterator = x + down;
      yiterator = y - arm;
      if (yiterator <= 0) {
        yiterator = 1;
        ++down;
        ++xiterator;
      }
      while (xiterator < n && yiterator < y) {
        if (xiterator == n - 1 && yiterator == 1) bottomRight = false;
        ++counter;
        ++xiterator;
        ++yiterator;
      }
    }
    if (topRight) {
      xiterator = x + arm;
      yiterator = y + right;
      if (xiterator >= n) {
        xiterator = n - 1;
        ++right;
        ++yiterator;
      }
      while (xiterator > x && yiterator < n) {
        if (xiterator == n - 1 && yiterator == n - 1) topRight = false;
        ++counter;
        --xiterator;
        ++yiterator;
      }
    }
    ++arm;
    ++timer;
  }
  std::cout << timer << endl;
  return 0;
}
