[CodeForces - 197B] B - Limit

B - Limit

You are given two polynomials:

• P(x) = a₀·xⁿ + a₁·x^n - 1 + ... + a_n - 1·x + a_n and
• Q(x) = b₀·x^m + b₁·x^m - 1 + ... + b_m - 1·x + b_m.

Calculate limit .

Input

The first line contains two space-separated integers n and m (0 ≤ n, m ≤ 100) — degrees of polynomials P(x) and Q(x) correspondingly.

The second line contains n + 1 space-separated integers — the factors of polynomial P(x): a₀, a₁, ..., a_n - 1, a_n ( - 100 ≤ a_i ≤ 100, a₀ ≠ 0).

The third line contains m + 1 space-separated integers — the factors of polynomial Q(x): b₀, b₁, ..., b_m - 1, b_m ( - 100 ≤ b_i ≤ 100, b₀ ≠ 0).

Output

If the limit equals  + ∞, print "Infinity" (without quotes). If the limit equals  - ∞, print "-Infinity" (without the quotes).

If the value of the limit equals zero, print "0/1" (without the quotes).

Otherwise, print an irreducible fraction — the value of limit , in the format "p/q" (without the quotes), where p is the — numerator, q (q > 0) is the denominator of the fraction.

Example

Input

2 11 1 12 5

Output

Infinity

Input

1 0-1 32

Output

-Infinity

Input

0 111 0

Output

0/1

Input

2 22 1 64 5 -7

Output

1/2

Input

1 19 0-5 2

Output

-9/5

Note

Let's consider all samples:

1. 
2. 
3. 
4. 
5. 

You can learn more about the definition and properties of limits if you follow the link: http://en.wikipedia.org/wiki/Limit_of_a_function

这个题目的意思就是,给你两个关于x的多项式,一个作为分子,一个作为分母,x的取值趋于正无穷,求这个分式的值趋于什么(0,inf,-inf或某个确定的值)

其中最高项系数不为0.

那么,我们发现,由于x趋于无穷大,那么,除了两个最高项,其他就不用考虑了.

如果n>m,则趋于无穷大,至于是正还是负,要看两个系数同号还是异号.

如果n<m,则值趋向与0.

如果n=m,则输出某数的分数形式,还要用gcd约分一下,也要注意符号.

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 ],b[];
 int read(){
     ,f=; char ch=getchar();
     '){if (ch=='-') f=-f; ch=getchar();}
     +ch-',ch=getchar();
     return x*f;
 }
 int gcd(int x,int y){
     ?x:gcd(y,x%y);
 }
 int main(){
     n=read(),m=read();
     ; i<=n; i++) a[i]=read();
     ; j<=m; j++) b[j]=read();
     if (n>m){
         ]*b[]>) printf("Infinity"); else
         ]*b[]<) printf("-Infinity");
     }else
     if (n<m){
         printf("0/1");
     }else{
         ],y=b[];
         ) x=-x; ) y=-y;
         int K=gcd(x,y);
         x/=K,y/=K;
         ]*b[]<) printf("-");
         printf("%d/%d",x,y);
     }
     ;
 }