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;

}