Description：

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. 

CodeForces - 255D_#include

.

在初始某一个位置有一个点这个点每一秒会向四个方向扩散，然后求第几秒的时候可以超过K个

找一个最小符合条件的，我们可以二分枚举答案来做，然后我们怎么求时间为 t 时细胞的总数呢  假设边界是无穷大的

我们先写出前四秒 1   5   13    25   41我们先假设第0秒为1，第一秒就是1+4*1，第二秒就是  5+4*2，第三秒是13+4*3，

第四秒是25+4*4  不难发现规律  a【t】=a【t-1】+4*t，然后求得公式就是2*t*t+2*t+1。

加上边界条件以后就是减去超出边界的

 假设到 某一边的距离为 x  超出就为 2 * (t - 1 - x) * (t - 1 - x + 1) / 2  +  t -  x 这个方向两边都超出了

再然后就是超出的部分有相交的部分 就是角上  假设点到达该角对应两边的距离为 x y 那超出部分为 (t-1-x-y)(t-1-x-y+1)/2

AC代码：

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*/

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stdlib.h>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#include<math.h>
using namespace std;
typedef long long ll;
typedef double ld;
const int INF = 0x3f3f3f3f;
const int N=5000;
int i,j,k;
int p,q;
int res,cnt,ans,temp;
ll n,x,y,m;
ll cul(ll p)
{
    if(p<=0)
        return 0;
    return
    (p+1)*p/2;
}
int judge(ll p)
{
    ll a=x-1,b=y-1,c=n-x,d=n-y;
    ll ans=min(p,a)+min(p,b)+min(p,c)+min(p,d)+1;
    p--;
    ans+=cul(p)*4-2*cul(p-a)-2*cul(p-b)-2*cul(p-c)-2*cul(p-d);
    ans+=cul(p-a-b)+cul(p-b-c)+cul(p-c-d)+cul(p-d-a);
    return ans>=m;
}
int main()
{
    scanf("%lld %lld %lld %lld",&n,&x,&y,&m);
    ll l=0,r=1e9,ans=0;
    while(l<=r)
    {
        ll mid=(r+l)>>1;
        if(judge(mid))
        {
            ans=mid;
            r=mid-1;
        }
        else l=mid+1;
    }
    printf("%lld\n",ans);
    return 0;
}