#include <bits/stdc++.h>

using namespace std;

const int maxn =  + ;

struct{

    int l, r, sum;

}tree[maxn*];//起码要*4

int a[maxn];

int n;

void build(int index, int l , int r){

    tree[index].l = l, tree[index].r = r;

    if( l == r){

        tree[index].sum = a[l];//递归建树，当l=r就是叶子节点没办法再分，直接赋值

        return;

    }

    int mid = (l+r) / ;

    build(index * , l, mid);

    build(index *  + , mid +  , r);

    tree[index].sum = tree[index * ].sum + tree[index* + ].sum;//不是叶子节点， 总和等于它儿子的总和

}

void update(int index, int a, int val){ //修改总和的

    if(tree[index].l == a && tree[index].r == a){

        tree[index].sum += val;//找到该点，直接更改

        return;

    }

    int mid = (tree[index].l + tree[index].r) / ;

    if(a <= mid)

        update(index * , a , val);

    else

        update(index *  + , a, val);

    tree[index].sum += val;//路径上的点都要更改。

}

int query(int index , int ql, int qr){

    if(tree[index].l == ql && tree[index].r == qr){

        return tree[index].sum;

    }

    int mid = (tree[index].l + tree[index].r) / ;

    if(qr <= mid) //查询区间只在这个的左子树

        return query(index * , ql, qr);

    if(ql > mid)  //只在右子树

        return query(index *  + , ql, qr);

    return query(index * , ql, mid) + query(index *  + , mid + , qr);//左右子树都有的话， 拆分查询区间

}

int main(){

    int T;

    scanf("%d", &T);

    int kase = ;

    while(T--){

        printf("Case %d:\n", kase++);

        scanf("%d", &n);

        for(int i = ; i <= n; i++){

            scanf("%d", &a[i]);

        }

        build(,,n);

        char cho[];

        while(scanf("%s",cho) && cho[] != 'E'){

            int x, y;

            scanf("%d %d", &x, &y);

            if(cho[] == 'Q'){

                printf("%d\n", query(,x,y));

            }

            else if(cho[] == 'A'){

                update(,x,y);

            }

            else update(,x,-y);

        }

    }

    return ;

}