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 cuadros(int n, int x, int y, int nivel) {
  int aux = x;
  x = y;
  y = aux;
  y = (n + 1) - y;
  long long sumaArr = 0;
  long long sumaAb = 0;
  int distIzq = x - 1;
  int distDer = n - x;
  int distArr = n - y;
  int distAb = y - 1;
  int cuadrosIzq = nivel;
  int cuadrosDer = nivel;
  if (distIzq < cuadrosIzq) cuadrosIzq = distIzq;
  if (distDer < cuadrosDer) cuadrosDer = distDer;
  for (int i = 0; i < cuadrosDer + 1; i++) {
    int c = nivel - i;
    if (distArr < c)
      sumaArr += 1 + distArr;
    else
      sumaArr += 1 + c;
    if (distAb < c)
      sumaAb += distAb;
    else {
      sumaAb += c;
    }
  }
  for (int i = 1; i < cuadrosIzq + 1; i++) {
    int c = nivel - i;
    if (distArr < c)
      sumaArr += 1 + distArr;
    else
      sumaArr += 1 + c;
    if (distAb < c)
      sumaAb += distAb;
    else {
      sumaAb += c;
    }
  }
  return sumaArr + sumaAb;
}
int main() {
  int n, x, y, limite;
  cin >> n >> x >> y >> limite;
  int izq = 0, der = 10000000, mid;
  while (izq < der) {
    mid = (izq + der) / 2;
    if (limite <= cuadros(n, x, y, mid))
      der = mid;
    else
      izq = mid + 1;
  }
  cout << der << endl;
  return 0;
}
