uva 11768

// 扩展欧几里得算法 
// 先求出一个解 再求出区间 [x1,x2]有几个整数符合条件
// 需要注意的是 水平和垂直2种情况的处理 还有正数和负数取整的细微差别
#include <iostream>
#include <algorithm>
#include <queue>
#include <stack>
#include <math.h>
#include <stdio.h>
#include <string.h>
using namespace std;
#define MOD 1000000007
#define maxn 1000010
#define maxm 48010
#define LL  long long
LL ax,ay,bx,by;
LL a,b,c;
LL d,x,y;
void extendGcd(LL a,LL b){
    if(b==0){
        d=a;
        x=1;
        y=0;
    }else{
       extendGcd(b,a%b);
       LL t=x;
       x=y;
       y=t-a/b*y;
    }
}
LL lt(double p){
    if(p>0){
        LL x=p+0.05;x=x*10;
        LL y=(p+0.05)*10;
        if(x==y) return x/10;
        return x/10+1;
    }
    else{
        return (LL)(p-0.05);
    }
}
LL rt(double p){
    if(p>=0)
      return (LL)(p+0.05);
    else {
        LL x=p-0.05;x=x*10;
        LL y=(p-0.05)*10;
        if(x==y) return x/10;
        return x/10-1;
    }
}
LL get(double d){
   if(d>=0) return (d+0.05)*10;
   else return (d-0.05)*10;
}
int main(){

  int T;
  double x1,y1,x2,y2;
  LL ax,ay,bx,by;
  LL a,b,c;
  scanf("%d",&T);
  while(T--){
    scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
    ax=get(x1);ay=get(y1);
    bx=get(x2);by=get(y2);
    if(ay==by){
        if(ay%10) { printf("0\n");continue;}
        LL lx=lt(min(x1,x2)),rx=rt(max(x1,x2));
        if(lx==rx&&ay%10){printf("0\n");continue;}
         printf("%lld\n",rx-lx+1);
        continue;
    }
    else if(ax==bx) {
        if(ax%10) { printf("0\n");continue;}
        LL ly=lt(min(y1,y2)),ry=rt(max(y1,y2));
        if(ly==ry&&ax%10){printf("0\n");continue;}
        printf("%lld\n",ry-ly+1);
        continue;
    }

    a=(ay-by)*10;
    b=(bx-ax)*10;
    c=(bx-ax)*ay-(by-ay)*ax;

    extendGcd(a,b);
    if(c%d!=0){printf("0\n");continue;}
    x=x*(c/d);
    LL m=b/d;
    LL lx=lt(min(x1,x2)),rx=rt(max(x1,x2));
    if(m<0) m=-m;
    x=x-(x-lx)/m*m;
    x-=m;
    while(x<lx)
        x+=m;
    LL k=(rx-x)/m;
    while(x+k*m<=rx) k++;
    printf("%lld\n",k);
  }
  return 0;
}