#include<bits/stdc++.h>
#define N 500005
#define int long long
#define db double
#define fo(i,l,r) for(int i=l;i<=r;i++)
#define rf(i,r,l) for(int i=r;i>=l;i--)
using namespace std;
int T,n,m,k,op,x,y,z,ans,last;
int a[N<<1],b[N<<1],c[N<<1],p[N],t[N];
void swap(int &x,int &y){int t=x;x=y;y=t;}
int max(int x,int y){return x>y?x:y;}
int min(int x,int y){return x<y?x:y;}
void ckmax(int &x,int y){if(y>x)x=y;}
void ckmin(int &x,int y){if(y<x)x=y;}
int fp(int a,int n,int mod){
    int ans=1;
    while(n){
        if(n&1)ans=ans*a%mod;
        a=a*a%mod;
        n>>=1;
    }
    return ans;
}
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int po(int x){return x*x;}
int sub(int x,int y){return x>y?x-y:y-x;}
int ab(int x){return x<0?-x:x;}
int read(){
    int sum=0;bool f=0;char c=getchar();
    for(;c<48||c>57;c=getchar())if(c==45)f=1;
    for(;c>=48&&c<=57;c=getchar())sum=sum*10+(c&15);
    return f?-sum:sum;
}
signed main(){
    freopen("cantrefuse.in","r",stdin);freopen("cantrefuse.out","w",stdout);
    n=read();m=read();
    fo(i,1,n)a[i]=read();
    fo(i,1,n)b[i]=read();
    fo(i,1,n-1)c[i]=read();
    fo(i,1,m)p[i]=read(),a[i+n]=read(),b[i+n]=read(),c[i+n]=read();
    bool f=1;
    fo(i,1,n+m)if(a[i]){f=0;break;}
    if(f){
        while(m--)printf("0\n");
        return 0;
    }
    f=1;
    fo(i,1,n+m)if(b[i]){f=0;break;}
    if(f){
        while(m--)printf("0\n");
        return 0;
    }
    f=1;
    fo(i,1,n+m)if(c[i]){f=0;break;}
    if(f){
        fo(i,1,n)ans+=min(a[i],b[i]);
        fo(i,1,m){
            ans-=min(a[p[i]],b[p[i]]);
            a[p[i]]=a[i+n];b[p[i]]=b[i+n];
            ans+=min(a[p[i]],b[p[i]]);
        }
        return 0;
    }
    fo(i,1,n){
        x=min(a[i]+t[i-1],b[i]);
        t[i]=min(a[i]+t[i-1]-x,c[i]);
        ans+=x;
    }
    fo(i,1,m){
        y=t[p[i]];
        ans-=min(a[p[i]]+t[p[i]-1],b[p[i]]);
        a[p[i]]=a[i+n];b[p[i]]=b[i+n];c[p[i]]=c[i+n];
        x=min(a[p[i]]+t[p[i]-1],b[p[i]]);
        t[p[i]]=min(a[p[i]]+t[p[i]-1]-x,c[p[i]]);
        ans+=x;
        fo(j,p[i]+1,n){
            if(t[j-1]==y)break;
            ans-=min(a[j]+y,b[j]);
            y=t[j];
            x=min(a[j]+t[j-1],b[j]);
            t[j]=min(a[j]+t[j-1]-x,c[j]);
            ans+=x;
        }
        printf("%lld\n",ans);
    }
    return 0;
}