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用c1存表示a[i]-a[i-1]的树状数组
用c2存(i-1)*c1[i]

sum[x]=c1[1]+c1[1]+c1[2]+c1[1]+c1[2]+c1[3]+...+c1[1]+...+c1[x]
          =x*(c1[1]+c1[2]+c1[3]+…+c1[x])-(0*c1[1]+1*c1[2]+2*c1[3]+…+(x-1)*c1[x])
#include<iostream>
#include<cstdio>
using namespace std;

const long long MAXN=1e6+5;

long long c1[MAXN],c2[MAXN];
long long n,m;

long long lowbit(long long x){return x&(-x);}
void upd(long long f,long long t,long long w);
long long ask(long long p);

int main(){
	scanf("%d%d",&n,&m);
	for(long long i=1;i<=n;++i){
		long long t;
		cin>>t;
		upd(i,i,t);
	}
	for(long long i=1;i<=m;++i){
		char q;
		cin>>q;
		if(q=='Q'){
			long long t1,t2;
			cin>>t1>>t2;
			long long ans=ask(t2)-ask(t1-1);
			cout<<ans<<'\n';
		}
		else{
			long long t1,t2,t3;
			cin>>t1>>t2>>t3;
			upd(t1,t2,t3);
		}
	}
	return 0;
}

void upd(long long f,long long t,long long w){
	long long add1=(f-1)*w,add2=t*w;
	for(long long i=f;i<=n;i+=lowbit(i)){
		c1[i]+=w;
		c2[i]+=add1;
	}
	for(long long i=t+1;i<=n;i+=lowbit(i)){
		c1[i]-=w;
		c2[i]-=add2;
	}
}

long long ask(long long p){
	long long ans1=0,ans2=0;
	for(long long i=p;i>0;i-=lowbit(i)){
		ans1+=c1[i];
		ans2+=c2[i];
	}
	return p*ans1-ans2;
}

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