#pragma GCC optimize("O3") #include #define ll long long #define rep(i,n) for(ll i=0;i<(n);i++) #define pll pair #define pii pair #define pq priority_queue #define pb push_back #define eb emplace_back #define fi first #define se second #define endl '\n' #define ios ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0); #define lb(c,x) distance(c.begin(),lower_bound(all(c),x)) #define ub(c,x) distance(c.begin(),upper_bound(all(c),x)) using namespace std; inline int topbit(unsigned long long x){ return x?63-__builtin_clzll(x):-1; } inline int popcount(unsigned long long x){ return __builtin_popcountll(x); } inline int parity(unsigned long long x){//popcount%2 return __builtin_parity(x); } template inline bool chmax(T& a,T b){if(a inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;} const ll INF=1e15; const ll mod=1e9+7; struct mint { ll x; // typedef long long ll; mint(ll x=0):x((x%mod+mod)%mod){} mint operator-() const { return mint(-x);} mint& operator+=(const mint a) { if ((x += a.x) >= mod) x -= mod; return *this; } mint& operator-=(const mint a) { if ((x += mod-a.x) >= mod) x -= mod; return *this; } mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;} mint operator+(const mint a) const { return mint(*this) += a;} mint operator-(const mint a) const { return mint(*this) -= a;} mint operator*(const mint a) const { return mint(*this) *= a;} mint pow(ll t) const { if (!t) return 1; mint a = pow(t>>1); a *= a; if (t&1) a *= *this; return a; } // for prime mod mint inv() const { return pow(mod-2);} mint& operator/=(const mint a) { return *this *= a.inv();} mint operator/(const mint a) const { return mint(*this) /= a;} }; istream& operator>>(istream& is, mint& a) { return is >> a.x;} ostream& operator<<(ostream& os, const mint& a) { return os << a.x;} // combination mod prime // https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619 struct combination { vector fact, ifact; combination(ll n):fact(n+1),ifact(n+1) { assert(n < mod); fact[0] = 1; for (ll i = 1; i <= n; ++i) fact[i] = fact[i-1]*i; ifact[n] = fact[n].inv(); for (ll i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i; } mint operator()(ll n, ll k) { if (k < 0 || k > n) return 0; return fact[n]*ifact[k]*ifact[n-k]; } mint p(ll n, ll k) { return fact[n]*ifact[n-k]; } } c(1000005); ll gcd(ll a, ll b) { return b?gcd(b,a%b):a;} int main(){ ios; ll n,m; cin >> n >> m; vector a(n),b(m); rep(i,n){ cin >> a[i]; } rep(i,m){ cin >> b[i]; } mint ans=0; rep(i,n){ rep(j,m){ if(a[i]==0) continue; ll g=gcd(a[i],b[j]); ll sum=b[j]*(b[j]+1)/2; a[i]/=g; b[j]/=g; ans+=(2*a[i]*sum-g*b[j]*(b[j]-1))/(b[j]); a[i]*=g; b[j]*=g; } } cout << ans << endl; return 0; }