C. Mike and Frog

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mike has a frog and a flower. His frog is named Xaniar and his flower is named Abol. Initially(at time 0), height of Xaniar is h1 and height of Abol is h2. Each second, Mike waters Abol and Xaniar.

So, if height of Xaniar is h1 and height of Abol is h2, after one second height of Xaniar will become  and height of Abol will become  where x1, y1, x2 and y2 are some integer numbers and  denotes the remainder of amodulo b.

Mike is a competitive programmer fan. He wants to know the minimum time it takes until height of Xania is a1 and height of Abol is a2.

Mike has asked you for your help. Calculate the minimum time or say it will never happen.

Input

The first line of input contains integer m (2 ≤ m ≤ 106).

The second line of input contains integers h1 and a1 (0 ≤ h1, a1 < m).

The third line of input contains integers x1 and y1 (0 ≤ x1, y1 < m).

The fourth line of input contains integers h2 and a2 (0 ≤ h2, a2 < m).

The fifth line of input contains integers x2 and y2 (0 ≤ x2, y2 < m).

It is guaranteed that h1 ≠ a1 and h2 ≠ a2.

Output

Print the minimum number of seconds until Xaniar reaches height a1 and Abol reaches height a2 or print -1 otherwise.

Sample test(s)

input

5 4 2 1 1 0 1 2 3

output

3

input

1023 1 2 1 0 1 2 1 1

output

-1

此题由于模m，可以很容易想到循环节，然而我开始确实想得太简单了。 模m的循环里主要有以下情况

1）模m里的循环节不经过a

2）模m的只经过一次a，然后在另一个循环节里循环。

3）模m存在一个经过a的循环节。

对于前两种情况，记录下第一次经过a的时间。对于第二种情况，记录下第一次经过a的时间以及a的循环节的时间长度。

然后分情况讨论即可。

#include 
         
          #include 
          
           #include 
           
            #include 
            
             #define LL long longusing namespace std;int vist[1000007];void gao(LL h,LL a,LL x,LL y,LL &t,LL &l,LL m){	t=l=0;	memset(vist,-1,sizeof(vist));	vist[h]=0;	while(true){//		cout<
             
              <
              
               -1){ l=t=-1; return ; } vist[h]=l; if(h==a){ break; } }// cout<<"YES"<
               
                tmp){ t=-1; break; } vist[h]=l; if(h==a) { break; } } if(t!=-1) l=l-tmp; else l=tmp;}LL gcd(LL a,LL b){ if(b==0) return a; return gcd(b,a%b);}int main(){ LL m,a1,h1,x1,y1,a2,h2,y2,x2,t1,l1,t2,l2;// while(scanf("%I64d",&m)!=EOF){ cin>>m; // scanf("%d%d%d%d",&h1,&a1,&x1,&y1); cin>>h1>>a1>>x1>>y1; gao(h1,a1,x1,y1,t1,l1,m); // scanf("%d%d%d%d",&h2,&a2,&x2,&y2); cin>>h2>>a2>>x2>>y2; gao(h2,a2,x2,y2,t2,l2,m);// printf("l1=%lld t1=%lld l2=%lld t2=%lld\n",l1,t1,l2,t2); if(t1==-1&&t2==-1){ if(l1==l2)//printf("%d\n",l1); cout<
                
                 <
                 
                  t2){ swap(t1,t2),swap(l1,l2); } // printf("l1=%lld t1=%lld l2=%lld t2=%lld\n",l1,t1,l2,t2); if(abs(t2-t1)%gcd(l2,l1)!=0) puts("-1"); else{ int k=0; while(true){ if((t2+k*l2-t1)%l1==0) break; k++; }// printf("%d\n",t2+k*l2); cout<
                  
                   <