#include<bits/stdc++.h>
using i64 = long long;
#ifdef usetemplate
template <class T>
T quick_power(T a, T b, auto mod) {
    if (b == 0) return 1 % mod;
    T kt = quick_power(a, b >> 1, mod);
    kt = kt * kt % mod;
    if (b & 1) kt = kt * a % mod;
    return kt;
}
template <class T>
T mul(T a, T b, auto mod) {
    if (b == 0) return 0;
    T kt = mul(a, b >> 1, mod);
    kt = (kt + kt) % mod;
    if (b & 1) kt = (kt + a) % mod;
    return kt;
}
#endif
constexpr int N = 5e5+1;
std::array<int,N*4> f;//下标为i的节点对应的区间的和
std::array<int, N> a;//原数组最多N个点
int n, m;

inline void buildtree(int k, int l, int r) {//标号为k的点 [l,r],l,r分别为区间左右端点
    if (l == r) {
        f[k] = a[l];
        return;
    }
    int m = (l + r) >> 1;//取中间值
    buildtree(k + k, l, m);//建左子树
    buildtree(k + k + 1, m + 1, r);//建右子树
    f[k] = f[k + k] + f[k + k + 1];
}
inline void add(int k, int l, int r, int x, int y) {//x一定包含在l,r之间的
    f[k] += y;//x包含在[l,r],如果x+=y,那么f_[l,r]+=y
    if (l == r) return;
    int m = (l + r) >> 1;
    //如果x在左边那就加左边,如果x在右边那就加右边
    if (x <= m) add(k + k, l, m, x, y); else add(k + k + 1, m + 1, r, x, y);

}
inline int calc(int k, int l, int r, int s, int t) {//下标为k的节点，[l,r]里第s和第t为的和求出来
    if (l == s and r == t) return f[k];
    int m = (l + r) >> 1;
    if (t <= m) return calc(k + k, l, m, s, t);//left
    else if (s > m) return calc(k + k + 1, m + 1, r, s, t);//right
    else return calc(k + k, l, m, s, m) + calc(k + k + 1, m + 1, r, m + 1, t);//left+right
}
signed main() {

    std::cin.tie(nullptr)->sync_with_stdio(false);
    std::cin >> n >> m;
    for (int i = 1; i <= n; ++i) std::cin >> a[i];
    //一开始把线段树上每个节点的和统计出来
    buildtree(1, 1, n);//标号为1的点对应的是[1,n]这个区间
    int t, x, y;
    for (int i = 1; i <= m; ++i) {
        std::cin >> t >> x >> y;
        if (t == 1) add(1, 1, n, x, y);//一号点;[1,n];点下标是x，改的值是y
        else std::cout << calc(1, 1, n, x, y) << '\n';
    }
    return 0;
}