The equation

There is an equation ax + by + c = 0. Given a,b,c,x1,x2,y1,y2 you must determine, how many integer roots of this equation are satisfy to the following conditions : x1<=x<=x2,   y1<=y<=y2. Integer root of this equation is a pair of integer numbers (x,y).

Input

Input contains integer numbers a,b,c,x1,x2,y1,y2 delimited by spaces and line breaks. All numbers are not greater than 10⁸ by absolute value。

Output

Write answer to the output.

Sample Input

1 1 -3
0 4
0 4

Sample Output

4

大意：给你a,b,c,x1,y1,x2,y2。让你求ax+by+c=0在区间[x1,x2],[y1,y2]的整数解个数。
考虑扩展欧几里得： 得到特解x0,y0后。设a=a/gcd,b=b/gcd。通解为x=x0+b*k,y=y0-a*k;
则 得到不等式 x1<=x+b*k<=x2,y1<=y-a*k<=y2。
解k的范围，取交集就好了。注意考虑向上取整与向下取整。
比如下界就应该是想上取整，上界应该向下取整。

#include <iostream>
#include <algorithm>
#define maxn 2025
using namespace std;
typedef long long ll;
ll ans=0;
ll L=-1e9,R=1e9;
ll ex_gcd(ll a,ll b,ll &x,ll &y)
{
    if(b==0)
    {
        x=1;
        y=0;
        return a;
    }
    ll ans=ex_gcd(b,a%b,x,y);
    ll temp=x;
    x=y;
    y=temp-(a/b)*x;
    return ans;
}
ll down(ll x,ll y)//向下取整
{   ll res=0;
        res=x/y;
    if(x<0||y<0)
    {
        if(x%y!=0)
        {
            res--;
        }
    }
    return res;
}
ll upper(ll x,ll y)//向上取整
{   ll res=x/y;
    if(x>0&&y>0||x<0&&y<0)
    {
        if(x%y!=0)
        {
            res++;
        }
    }
    return res;
}
void sovle(ll l,ll r,ll b)
{
   if(b<0)
   {
       l=-l;
       r=-r;
       b=-b;
       swap(l,r);
   }
   L=max(L,upper(l,b));
   R=min(R,down(r,b));

}
int main()
{
    int n;
    ll a,b,c,x1,y1,x2,y2;
    cin>>a>>b>>c>>x1>>x2>>y1>>y2;
    c=-c;
    if(a==0&&b==0&&c!=0)
    {
        ans=0;
    }
    else if(a==0&&b==0&&c==0)
    {
        ans=(x2-x1+1)*(y2-y1+1);
    }
    else if(a==0&&b!=0)
    {
        if(c%b==0)
        {
         ll  y=c/b;
            if(y>=y1&&y<=y2)
            {
                ans++;
            }
        }
    }
    else if(b==0&&a!=0)
    {
        if(c%a==0)
        {
            ll x=c/a;
            if(x>=x1&&x<=x2)
            {
                ans++;
            }
        }
    }
    else
    {   ll x,y;
        ll gcd=ex_gcd(a,b,x,y);
        if(c%gcd==0)
        {   x=x*c/gcd;
            y=y*c/gcd;
            a/=gcd;
            b/=gcd;
            ll lx=x1-x,rx=x2-x;
            ll ly=y1-y,ry=y2-y;
           sovle(lx,rx,b);
           sovle(ly,ry,-a);
           ans=R-L+1;
        }
    }
    cout<<ans<<endl;
    return 0;
}