题目描述：

Vasya is reading a e-book. The file of the book consists of nn pages, numbered from 11 to nn. The screen is currently displaying the contents of page xx, and Vasya wants to read the page yy. There are two buttons on the book which allow Vasya to scroll dd pages forwards or backwards (but he cannot scroll outside the book). For example, if the book consists of 1010 pages, and d=3d=3, then from the first page Vasya can scroll to the first or to the fourth page by pressing one of the buttons; from the second page — to the first or to the fifth; from the sixth page — to the third or to the ninth; from the eighth — to the fifth or to the tenth.

Help Vasya to calculate the minimum number of times he needs to press a button to move to page yy.

Input

The first line contains one integer tt (1≤t≤1031≤t≤103) — the number of testcases.

Each testcase is denoted by a line containing four integers nn, xx, yy, dd (1≤n,d≤1091≤n,d≤109, 1≤x,y≤n1≤x,y≤n) — the number of pages, the starting page, the desired page, and the number of pages scrolled by pressing one button, respectively.

Output

Print one line for each test.

If Vasya can move from page xx to page yy, print the minimum number of times he needs to press a button to do it. Otherwise print −1−1.

INPUT：

3
10 4 5 2
5 1 3 4
20 4 19 3

OUTPUT：

4
-1
5

记得fmax和fmin要开G++编译器！

#include
#include
#include
using namespace std;
int main(){
int t,p,a,b,x;
cin>>t;
while(t--){
long long sum=0;
cin>>p>>a>>b>>x;
if(a==b){
cout<<0<<endl;
}else{
int k=abs(a-b);

        if(k%x==0){  
            sum=k/x;  
            cout<<sum<<endl;  
        }else{  
            /\*两种情况，a->1->b,a->p->b\*/  
            int m1,m2;  
            if((a-1)%x==0){  
                sum+=(a-1)/x;  
            }else{  
                sum+=(a-1)/x+1;  
            }  
            if((b-1)%x==0){  
                sum+=(b-1)/x;  
            }else{  
                sum=-1;  
            }  
            m1=sum;  
            sum=0;  
            if((p-a)%x==0){  
                sum+=(p-a)/x;  
            }else{  
                sum+=(p-a)/x+1;  
            }  
            if((p-b)%x==0){  
                sum+=(p-b)/x;  
            }else{  
                sum=-1;  
            }  
            m2=sum;  
            //cout<<m1<<endl<<m2<<endl<<endl;  
            if(m1!=-1&&m2!=-1){  
                if(m1<m2){  
                    cout<<m1<<endl;  
                }else{  
                    cout<<m2<<endl;  
                }  
                //cout<<fmin(m1,m2)<<endl;  
            }if(m1!=-1&&m2==-1){  
                cout<<m1<<endl;  
            }if(m1==-1&&m2!=-1){  
                cout<<m2<<endl;  
            }if(m1==-1&&m2==-1){  
                cout<<-1<<endl;  
            }  
        }  
    }  
}  
return 0;  

}