A - Ada and List

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AdatheLadybug有一份包含数字的待办事项列表。她需要在列表中进行插入、删除和查询操作。本篇介绍了一种使用C++ STL中的deque来高效处理这些操作的方法。

Ada the Ladybug has a TODO-list (containing only numbers - for simplicity). She is still doing something, so she sometimes erases k^th number, sometimes she inserts something on k^th position and sometime she asks for k^th number.

Sadly, she is now searching for a work at position k so she doesn't have time to do this herself. Can you help her?

Input

The first line will contain 0 < N ≤ 10⁵,0 < Q < 5*10⁵, the number of elements in TODO-list and number of queries.

Then a line with N numbers follows. Each number 0 ≤ A_k ≤ 10⁹ means k^th number in her TODO-list.

Afterward, Q lines follow, each beginning with number 1 ≤ a ≤ 3

1 k x means that you will add number x to position k

2 k means that you will erase number from position k

3 k means that you will print number from position k

For all queries, it is true that 1 ≤ k ≤ #SizeOfList, 0 ≤ x ≤ 10⁹ (for query 1, it can be also put to position #SizeOfList + 1)

You will never get query of type 2 or 3 if the list is empty

Output

For each query of type 3, print k^th numbers

Example Input

6 10
1 2 4 8 16 32
3 4
1 1 7
3 2
2 2
2 2
3 2
1 6 666
3 6
2 1
3 1

Example Output

Queries explanations

1 2 4 8 16 32
7 1 2 4 8 16 32
7 1 2 4 8 16 32
7 2 4 8 16 32
7 4 8 16 32
7 4 8 16 32
7 4 8 16 32 666
7 4 8 16 32 666
4 8 16 32 666
4 8 16 32 666

解析：
STL模拟；

#include<bits/stdc++.h>
using namespace std;
deque<int>q;
int n,Q;
int  num;
int a;
int k;
int x;

int main()
{
    scanf("%d%d",&n,&Q);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&num);
        q.push_back(num);
    }
    while(Q--)
    {
        scanf("%d",&a);
        if(a==1)
        {
            scanf("%d%d",&k,&x);
            ///long long tem=q.at(k-1)+x;
            ///cout<<tem<<endl;
            q.insert(q.begin()+k-1,x);
            ///q.erase(q.begin()+k);

        }
        else if(a==2)
        {
            scanf("%d",&k);
            q.erase(q.begin()+k-1);
        }
        else {
            scanf("%d",&k);
            printf("%d\n",q.at(k-1));
        }
    }
}