SQRT DECOMP

#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 nl "\n"
#define pp pair < ll , ll >
#define sz(x) (ll)x.size()
#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 = 2e5 + 1;
const int mod = 1e9 + 7;

// square root decomposition
ll a[N],  bsize ;
ll block[N];
// bsize is the size of block
void pre(ll n){
    //we can find the block number using i/bsize
    for(int i = 0 ; i < n; i++ ){
        block[i/bsize] += a[i];
    }
}
void update(ll idx , ll val){
    int id = idx/bsize;
    block[id] += val - a[idx];
    a[idx] = val;
}

ll query(ll l ,ll r){
    ll sum = 0 ;
    //for query there are three points
    //1.find the sum of block which starts from l
    //we are taking l < r because in last loop we are considring the case l <= r
    while(l % bsize != 0 && l < r){
        sum += a[l];
        l++;
    }
    //2.find the sum of all the block b/w l and r
    //take the case l starts from 1 and blk size is 3 then l + p -1 will points to 3
    //which is second block
    while(l + bsize - 1 <= r){
        sum += block[l/bsize];
        l += bsize;
    }
    //3.find the sum of block where r ends

    while(l <= r){
        sum += a[l];
        l++;
    }
    return sum;

}
//time complexity of query will be O(sqrt(n));

int main()
{
    fast;
    ll n;
    cin >> n ;
    bsize = (sqrt(n));
   //   cout << bsize << " ";

    for(int i = 0 ; i < n; i++ )
        cin >> a[i];
    pre(n);

    ll q;
    cin >> q;
    while(q--){
        ll ch;
        cin >> ch ;
        if(ch == 1){
            ll idx , val;
            cin >> idx >> val;
            idx--;
            update(idx ,val);
        }
        else{
            ll a , b;
            cin >> a >> b ;
            a-- , b--;
            cout << query(a,b) << "\n";
        }
    }






















    return 0;

}