#include<bits/stdc++.h>

using namespace std;

const double  Pi = acos(-1.0);

struct cplex
{
    double  r,i;
    cplex(double  _r = 0.0 , double  _i = 0.0)
    {
        r = _r , i = _i;
    }
    cplex operator+(const cplex &b)
    {
        return cplex(r+b.r,i+b.i);
    }
    cplex operator-(const cplex &b)
    {
        return cplex(r-b.r,i-b.i);
    }
    cplex operator * (const cplex &b)
    {
        return cplex(r*b.r-i*b.i,r*b.i+i*b.r);
    }
};
void change(vector<cplex> &y)
{
    int ln=y.size();
    for(int i=1,j=ln>>1; i<ln-1; i++)
    {
        if(i<j)swap(y[i],y[j]);
        int k=ln>>1;
        while(j>=k)
        {
            j-=k;
            k>>=1;
        }
        j+=k;
    }
}
void fft(vector<cplex> &y,int on)
{
    change(y);
    int len=y.size();
    for(int m=2; m<=len; m<<=1)
    {
        cplex wm(cos(-on*2*Pi/m),sin(-on*2*Pi/m));
        for(int k=0; k<len; k+=m)
        {
            cplex w(1,0);
            for(int j=0; j<m/2; j++)
            {
                cplex u=y[k + j];
                cplex t=w*y[k+j+m/2];
                y[k+j]=u+t;
                y[k+j+m/2]=u-t;
                w=w*wm;
            }
        }
    }
    if(on==-1)for(int i=0; i<len; i++)y[i].r/=len;
}
typedef long long ll;
vector<ll> mult(vector<ll> a,vector<ll> b)
{
    int len=1;
    int la=a.size();
    int lb=b.size();
    while(len<la+lb) len<<=1;
    vector<cplex> x1(len);
    vector<cplex> x2(len);
    for(int i=0; i<la; i++) x1[i]= cplex(a[i],0);
    for(int i=la; i<len; i++)x1[i] =cplex(0,0);
    for(int i=0; i<lb; i++)x2[i]= cplex(b[i],0);
    for(int i=lb; i<len; i++)x2[i]= cplex(0,0);
    fft(x1,1);
    fft(x2,1);
    for(int i = 0 ; i < len ; i++) x1[i] = x1[i] * x2[i];
    fft(x1,-1);
    vector<ll> sol(len);
    for(int i=0; i<len; i++)
    {
        sol[i]=x1[i].r+0.5;
    }
    return sol;
}
const int SQ = 1314, MX = 200000;
int T , n , QN;
int A[MX] , B[MX];
vector < ll > results[MX/SQ];
void solve(){

    scanf("%d %d",&n,&QN);

    assert(1 <= n && n <= 100000);

    for(int j = 1 ; j <= n ; j++) scanf("%d",&A[j]);
    for(int j = 1 ; j <= n ; j++) scanf("%d",&B[j]);

    vector < ll > values;
    for(int j = 1 ; j <= n ; j++) values.push_back(A[j]);


    for(int a = 1 ; a <= n ; a+=SQ){

        int idx = a / SQ;

        int b = a + SQ - 1;

        b = min(b , n);

        vector < ll > v;
        for(int j = a ; j <= b ; j++)
            v.push_back(B[j]);

        reverse(v.begin() , v.end());

        results[idx] = mult(values , v);

    }

    while(QN--){
        int l1 , l2 , r1 , r2 , len;
        scanf("%d %d %d",&l1,&l2,&len);
        r1 = l1 + len - 1;
        r2 = l2 + len - 1;

        ll ans = 0;

        for(int a = ( (l2 - 1) / SQ )*SQ + 1 ; a <= r2 ; a += SQ){

            int idx = a / SQ;

            int b = a + SQ - 1;

            b = min(b , n);

            if(l2 > b || r2 < a) continue;

            //cout<<a<<' '<<b<<endl;

            if(a >= l2 && b <= r2){

                int upstart = l1 + a - l2;
                int inmulindex = min(SQ , n - a + 1);
                int sumindex = upstart - 1 + inmulindex - 1;

                ans += results[idx][sumindex];


            }

            else{
                int stindex = max(l2 , a) , enindex = min(r2 , b);

                for(int j = stindex ; j <= enindex ; j++){
                    ans += 1ll * B[j] * A[l1 + j - l2];
                }
            }
            //cout<<ans<<endl;

        }
        printf("%lld\n",ans);
    }

}
int main(){
    scanf("%d",&T);
    while(T--) solve();


}