ABC058 井井井/###

• 以前解いた時と解法が全く違ったので記念に。
• 前回は解説見て実装したからそれはそう。
• そう思うと解説のような綺麗な解き方するのは中々難しい。。

#include<bits/stdc++.h>
#define fi first
#define se second
#define show(x) cout<<#x<<"="<<x<<"\n"
typedef long long ll;
template<typename T>
void print_to(std::ostream &os,const char *,const char *tail,const T &fst){
  os<<fst<<tail;
}
template<typename Fst,typename... Rst>
void print_to(std::ostream &os,const char *del,const char *tail,const Fst &fst,const Rst &... rst){
  os<<fst<<del;
  print_to(os,del,tail,rst...);
}
template<typename Iter>
void print_to_(std::ostream &os,const char *del,const char *tail,Iter bgn,Iter end){
  for(Iter it=bgn;it!=end;){
    os<<*it;
    if(++it!=end)os<<del;
    else os<<tail;
  }
}
template<typename Fst,typename... Rst>
void println(const Fst &fst,const Rst &... rst){
  print_to(std::cout,"\n","\n",fst,rst...);
}
template<typename Fst,typename... Rst>
void print(const Fst &fst,const Rst &... rst){
  print_to(std::cout," ","\n",fst,rst...);
}
template<typename Iter>
void println_(Iter bgn,Iter end){
  print_to_(std::cout," ","\n",bgn,end);
}
using namespace std;
const ll MOD=(ll)1e9+7;
const int dy[]={1,0,-1};
const int dx[]={1,0,-1};
int n,m,h,w;
string s;
ll x[100001],y[100001],sumy[100001],sumx[100001];


int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(10);
  cout<<fixed;
#ifdef LOCAL_DEFINE
  FILE *stream1;
  //FILE *stream2;
  stream1=freopen("in","r",stdin);
  //stream2=freopen("out","w",stdout);
  if(stream1==NULL)return 0;
  //if(stream2==NULL)return 0;
#endif
  cin>>n>>m;
  for(int i=0;i<n;i++){
    cin>>x[i];
  }
  for(int i=0;i<m;i++){
    cin>>y[i];
  }
  for(int i=m-1;i>=0;i--){
    if(i==m-1){
      sumy[i]=y[i];
      continue;
    }
    sumy[i]=y[i]+sumy[i+1];
  }
  for(int i=n-1;i>=0;i--){
    if(i==n-1){
      sumx[i]=x[i];
      continue;
    }
    sumx[i]=x[i]+sumx[i+1];
  }
  ll sumyy=0;
  for(int i=0;i<m;i++){
    ll p=i;
    sumyy=(sumyy+MOD)%MOD+(sumy[p+1]+MOD)%MOD+-(((m-1-p)%MOD)*(y[p]%MOD)+MOD)%MOD;
    sumyy%=MOD;
    //print(sumyy);
  }
  ll sumxx=0;
  for(int i=0;i<n;i++){
    ll p=i;
    sumxx=(sumxx+MOD)%MOD+(sumx[p+1]+MOD)%MOD-(((n-1-p)%MOD)*(x[p]%MOD)+MOD)%MOD;
    sumxx%=MOD;
  }
  //show(sumxx);
  //show(sumyy);
  print((((sumxx+MOD)%MOD)*((sumyy+MOD)%MOD)+MOD)%MOD);
#ifdef LOCAL_DEFINE
  cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<"s.\n";
  fclose(stream1);
  //fclose(stream2);
#endif
    return 0;
}