/*
De bai:
Cho mảng A có N số và Q truy vấn:
 
Có hai loại truy vấn:
1 pos val: chỉnh a[pos] += val
2 l r: Tính tổng từ l đến r
*/
 
#include <bits/stdc++.h>
#define endl '\n'
typedef long long ll;
using namespace std;
 
const ll MAXN = 1e5;
ll f[MAXN], n, q;
 
void update(ll pos, ll val) { // Cap nhat vi tri pos len val don vi
    for (ll i = pos; i <= n; i += i & -i) f[i] += val;
    // Em khong can hieu i & -i la gi dau, chi can hoc thuoc
}
 
ll get(ll pos) { // Ham get nay se tra ve tong tu 1 den pos
    ll res = 0;
    for (ll i = pos; i; i -= i & -i) res += f[i];
    return res;
    /*
    Vi du em muon tinh tong tu l den r thi giong prefix sum: get(r) - get(l - 1)
    */
}
 
void sol() {
    cin >> n >> q;
    for (ll i = 1; i <= n; i++) {
        ll x; cin >> x;
        update(i, x);
    }
    for (ll i = 1; i <= q; i++) {
        ll t, u, v; cin >> t >> u >> v;
        if (t == 1) update(u, v);
        else cout << get(v) - get(u - 1) << endl;
    }
}
 
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
 
//    freopen(".INP", "r", stdin);
//    freopen(".OUT", "w", stdout);
 
    sol();
 
    return 0;
}
 

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NCAzCjEgMiAzIDQKMiAyIDQKMSAyIDEKMiAyIDQKOQoxMA==

4 3
1 2 3 4
2 2 4
1 2 1
2 2 4
9
10