codeforces 315 B.Sereja and Array(线段树区间更新+单点更新+单点询问)

Sereja has got an array, consisting of n integers, a₁, a₂, ..., a_n. Sereja is an active boy, so he is now going to complete m operations. Each operation will have one of the three forms:

1. Make v_i-th array element equal to x_i. In other words, perform the assignment a_vi = x_i.
2. Increase each array element by y_i. In other words, perform n assignments a_i = a_i + y_i (1 ≤ i ≤ n).
3. Take a piece of paper and write out the q_i-th array element. That is, the element a_qi.

Help Sereja, complete all his operations.

Input

The first line contains integers n, m (1 ≤ n, m ≤ 10⁵). The second line contains nspace-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹) — the original array.

Next m lines describe operations, the i-th line describes the i-th operation. The first number in the i-th line is integer t_i (1 ≤ t_i ≤ 3) that represents the operation type. If t_i = 1, then it is followed by two integers v_i and x_i, (1 ≤ v_i ≤ n, 1 ≤ x_i ≤ 10⁹). If t_i = 2, then it is followed by integer y_i (1 ≤ y_i ≤ 10⁴). And if t_i = 3, then it is followed by integer q_i (1 ≤ q_i ≤ n).

Output

For each third type operation print value a_qi. Print the values in the order, in which the corresponding queries follow in the input.

Example

Input

10 111 2 3 4 5 6 7 8 9 103 23 92 103 13 101 1 102 102 103 13 103 9

Output

291120304039

题解：

题意：

操作1 x y表示将a[x]=y

操作2 x 表示全体数列值加上x

操作3 x 表示询问a[x]处的值

思路：

比较裸的一道线段树的题，然后这里的区间更新因为是全体更新所以可以偷懒

代码：

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<stack>#include<math.h>#include<vector>#include<map>#include<set>#include<stdlib.h>#include<cmath>#include<string>#include<algorithm>#include<iostream>using namespace std;#define lson k*2#define rson k*2+1#define M (t[k].l+t[k].r)/2struct node{    int v;    int tag;    int l,r;}t[100005*4];void pushdown(int k){    if(t[k].tag)    {        t[lson].tag+=t[k].tag;        t[rson].tag+=t[k].tag;        if(t[lson].l==t[lson].r)            t[lson].v+=t[k].tag;        if(t[rson].l==t[rson].r)            t[rson].v+=t[k].tag;        t[k].tag=0;    }}void Build(int l,int r,int k){    t[k].l=l;    t[k].r=r;    t[k].tag=0;    if(l==r)    {        scanf("%d",&t[k].v);        return;    }    int mid=M;    Build(l,mid,lson);    Build(mid+1,r,rson);}void update(int k,int v){    t[k].v+=v;    t[k].tag+=v;}void change(int x,int v,int k){    if(t[k].l==x&&t[k].r==x)    {        t[k].v=v;        return;    }    pushdown(k);    int mid=M;    if(x<=mid)        change(x,v,lson);    else        change(x,v,rson);}int query(int x,int k){    if(t[k].l==x&&t[k].r==x)    {        return t[k].v;    }    pushdown(k);    int mid=M;    if(x<=mid)        return query(x,lson);    else        return query(x,rson);}int main(){    int i,j,n,m,d,x,y;    scanf("%d%d",&n,&m);    Build(1,n,1);    while(m--)    {        scanf("%d",&d);        if(d==1)        {            scanf("%d%d",&x,&y);            change(x,y,1);        }        else if(d==2)        {            scanf("%d",&x);            update(1,x);        }        else        {            scanf("%d",&x);            printf("%d\n",query(x,1));        }    }    return 0;}