SEG TREE SUM RANGE SINGLE UPDATE

    /*If you are slow ,try hard and become fast
        If you are making mistakes , learn from them and correct them
            If you are weak in something , do practice and become good at it
    */
/* Don't do just for the sake ... learn from it , that is more important */

//Don't Think Just Do It!!

#include <bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)
#define ll long long
#define ld long double
#define pb push_back
#define fe first
#define se second
#define st(x) sort(x.begin(),x.end())
#define rst(x) sort(x.rbegin(), x.rend())
#define all(x) x.begin(),x.end()
long double pi = 3.14159265358979323;

const double EPS = 1e-12;
const int N = 1e5 + 5;
const int mod = 1e9 + 7;


// segment tree
ll tree[N << 2];
ll a[N];
void build(ll l , ll r , ll node)
{
    if(l == r){

        tree[node] = a[l];

        return ;
    }
    ll mid = ( l + r) >> 1;

    build(l,mid,node*2);
    build(mid + 1 , r , node*2 + 1);
    tree[node] = tree[node*2] + tree[node*2 + 1];

}

ll query(ll l , ll r ,ll node,ll ql , ll qr )
{
    if(r < ql || l > qr)//if queury is out of the range...
        return 0 ;
    if(l >= ql && r <= qr)//important condition like if our array is of size of 8 and we are
                         //quering for 1 - 6 then in second iteration for block 1 - 4;
         return tree[node];// // range represented by a node is completely inside the given range
    ll mid = (l + r) >> 1;
    ll s1 = query(l , mid ,node << 1, ql , qr );
    ll s2 = query(mid + 1 , r ,(node << 1) + 1,ql , qr );
    //learn very important note do NOT use node << 1 + 1 it will result in (node << 2)

    return s1 + s2;


}

void update(ll l , ll r ,ll node,ll idx,ll val)
{

    if(l == r){
        if(l == idx){
            tree[node] += val;
            a[l] += val;
        }
        return ;
    }
    else{
    ll mid = (l + r) >> 1;
    if(idx <= mid)
    update(l , mid  ,node << 1,idx,val);
    else
    update(mid + 1 , r ,(node << 1)  + 1,idx,val);
    tree[node] = tree[node << 1] + tree[(node << 1) + 1];
}

}
int main()
{
    fast;
    ll n;
    cin >> n;

    for(int i = 1 ; i <= n;  i++ )
        cin >> a[i];
    build(1,n,1);

    ll q;
    cin >> q;

    while(q--)
    {
        ll c;
        cin >> c;
        if(!c){
            ll ql , qr;
            cin >> ql >> qr;

            cout << query(1,n,1,ql,qr) << "\n";
        }
        else
        {
            ll idx,val;
            cin >> idx >> val;
            update(1,n,1,idx,val);
        }
    }



//time complexity
//build O(n)
//query O(logn)
//update O(logn)
//space complexity O(n)




















    return 0;

}