模拟，前缀和，排序，二分（或者尺取）

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
const ll mod=998244353;
const ll size=2e3+10;
ll n,m,ans,num,tb,ta,a[size],b[size],sa[size*size/2],sb[size*size/2];
inline ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return x*f;
}
ll gcd(ll a,ll b)
{
	if(b)while((a%=b)&&(b%=a));

	return a+b;
}
int main(){
    n=read();m=read();
   // num=n*(n+1)/2%mod;num=num*m%mod*(m+1)%mod;num=num*power(2,mod-2)%mod;
    for(ll i=1;i<=n;i++) a[i]=read(),a[i]+=a[i-1];
    for(ll i=1;i<=m;i++) b[i]=read(),b[i]+=b[i-1];
    for(ll l=1;l<=m;l++){
        for(ll r=l;r<=m;r++){
            sb[++tb]=b[r]-b[l-1];
        }
    }
    for(ll l=1;l<=n;l++){
        for(ll r=l;r<=n;r++){
            sa[++ta]=a[r]-a[l-1];
        }
    }
    sort(sa+1,sa+1+ta);
    sort(sb+1,sb+1+tb);
    ll l=0;
    for(ll i=1;i<=ta;i++){
        while(sb[l+1]<sa[i]&&l+1<=tb) l++;
        ans=ans+l;
    }
    int kk=gcd(ans,ta*tb);
	cout << ans/kk << "/" << ta*tb/kk;
    return 0;
}