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?

The first line contains four space-separated integers n, x, y, c(1 ≤ n, c ≤ 10⁹; 1 ≤ x, y ≤ n; c ≤ n²).

In a single line print a single integer — the answer to the problem.

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. .

//这题就是先算总面积，再减去超出面积，再加上重叠面积。关键要注意细节。
#include<iostream>
using namespace std;
int main()
{
     long long x,n,y,c;
     cin>>n>>y>>x>>c;
     long long i,xr,xl,yu,yd,d;
     x--;y--;
     long now=0;
     for(i=1;now<c;i++)
     {
         now=i*i+(i-1)*(i-1);
         if(now<c)
         continue;
         i--;
         xr=x+i;
         xl=x-i;
         yu=y-i;
         yd=y+i;
         if(xl<0) now-=xl*xl;
         if(xr>n-1) now-=(xr-n+1)*(xr-n+1);
          if(yu<0)
          {
              now-=yu*yu;
              yu++;
              d-=yu;
              if(x+d>n-1)
                now+=(x+d-n+2)*(x+d-n+1)/2;
              if(x+yu<0)
                now+=(x-d)*(x-d-1)/2;
          }
          if(yd>n-1)
          {
               now-=(yd-n+1)*(yd-n+1);
               d=yd-n+1;
               d--;
               if(x+d>n-1) now+=(x+d-n+2)*(x+d-n+1)/2;
               if(x-d<0) now+=(x-d)*(x-d-1)/2;
          }
          i++;
     }
     cout<<i-2<<endl;
}