（STL之vector）spoj-Ada and List（易）

Ada the Ladybug has a TODO-list (containing only numbers - for simplicity). She is still doing something, so she sometimes erases kth number, sometimes she inserts something on kth position and sometime she asks for kth 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 ≤ Ak ≤ 109 means kth 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 kth 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

8
1
4
666
4

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

使用vector就可以水过……

记住insert函数！

#include <iostream>
#include <string>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <stack>
#define mp make_pair
typedef long long ll;
typedef unsigned long long ull;
const int MAX=1e6+5;
const int INF=-1e9;
using namespace std;
typedef pair<int,int> pii;
int n,q;
int a,b,k;
int main()
{
    scanf("%d%d",&n,&q);
    vector <int> x;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a);x.push_back(a);
    }
    for(int i=1;i<=q;i++)
    {
        scanf("%d%d",&a,&b);
        if(a==1)
        {
            scanf("%d",&k);
            x.insert(x.begin()+b-1,k);
        }
        else if(a==2)
        {
            x.erase(x.begin()+b-1);
        }
        else
        {
            printf("%d
",x[b-1]);
        }
    }
}