/// solutie cu AINT ( arbori de intervale )

#include <bits/stdc++.h>

using namespace std;
ifstream f("datorii.in");
ofstream g("datorii.out");
vector<int> aint;
int n,m,c,st,dr,p,suma(int,int,int);
void upd(int,int);


int main()
{
    f>>n>>m;
    for(p=1;p<n;p*=2);
    aint = vector<int>(2*p,0);
    for(int i=p;i<p+n;i++)
        f>>aint[i];
    for(int i=p-1;i>=1;i--)
        aint[i]=aint[2*i]+aint[2*i+1];
    for(;m;m--)
    {
        f>>c>>st>>dr;
        if(c==0)
            upd(st,-dr);
        else
            g<<suma(1,1,p)<<'\n';
    }
    return 0;
}
void upd(int i,int x)
{
    i+=p-1;
    aint[i]+=x;
    for(i/=2;i;i/=2)
        aint[i]=aint[2*i]+aint[2*i+1];
}
int suma(int i,int lo,int hi)
{
    if(st>hi||lo>dr)return 0;
    if(st<=lo&&hi<=dr)return aint[i];
    int mi=(lo+hi)/2;
    return suma(2*i,lo,mi)+suma(2*i+1,mi+1,hi);
}