CF#344 div2 C.Report(思维)

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Each month Blake gets the report containing main economic indicators of the company "Blake Technologies". There are n commodities produced by the company. For each of them there is exactly one integer in the final report, that denotes corresponding revenue. Before the report gets to Blake, it passes through the hands of m managers. Each of them may reorder the elements in some order. Namely, the i-th manager either sorts first ri numbers in non-descending or non-ascending order and then passes the report to the manager i+1, or directly to Blake (if this manager has number i=m).

Employees of the "Blake Technologies" are preparing the report right now. You know the initial sequence ai of length n and the description of each manager, that is value ri and his favourite order. You are asked to speed up the process and determine how the final report will look like.

Input

The first line of the input contains two integers n and m (1≤n,m≤200000)− the number of commodities in the report and the number of managers, respectively.

The second line contains n integers ai (|ai|≤109)− the initial report before it gets to the first manager.

Then follow m lines with the descriptions of the operations managers are going to perform. The i-th of these lines contains two integers ti and ri (, 1≤ri≤n), meaning that the i-th manager sorts the first ri numbers either in the non-descending (if ti=1) or non-ascending (if ti=2) order.

Output

Print n integers− the final report, which will be passed to Blake by manager number m.

Examples

Input

3 1
1 2 3
2 2

Output

2 1 3

Input

Output

2 4 1 3

Note

In the first sample, the initial report looked like: 1 2 3. After the first manager the first two numbers were transposed: 2 1 3. The report got to Blake in this form.

In the second sample the original report was like this: 1 2 4 3. After the first manager the report changed to: 4 2 1 3. After the second manager the report changed to: 2 4 1 3. This report was handed over to Blake.

题目大意：给出n个数，m个操作。输入a,b，a为操作种类，b为操作位置。2种操作——1.1到b从小到大排序，2.从1到b从大到小排序。最后输出最终操作后的数组。

思路：直接模拟是不可能过的，妥妥的超时。所以需要思考如何缩减操作，若操作位置大于上一个操作的位置，那之前的操作都会被覆盖掉，所以可以从这一点入手。标记每一个操作的时间线与操作位置。可以考虑用2个数组存。例如：

第一个输出的数组是操作的时间线，第二个输出的是操作类型，第三个输出的是最终操作后的数组。可以看出，最后时间线最大的会覆盖掉之前操作位置比它小的操作。

代码如下：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=200010;
int p[maxn];//数组 
int num[maxn];//数组 
int s1[maxn];//操作时间  
int s2[maxn];//操作类型 
int main()
{
    int n,m;cin>>n>>m;
    for(int i=1;i<=n;++i)
	{
        scanf("%d",&num[i]);
    }
    int a,b;
    for(int i=1;i<=m;i++) 
	{
        scanf("%d%d",&a,&b);
        s1[b]=i;s2[b]=a;//s1记录对当前的数的操作的最晚的一次s2记录操作的类型
    }
    for(int i=1;i<=n;i++) printf("%d ",s1[i]);printf("\n");
	for(int i=1;i<=n;i++) printf("%d ",s2[i]);printf("\n"); 
    int F=1,R=n,nowoper,now=0;
    for(int i=n;i>=1;i--)
	{
        if(s2[i]==0)  p[i]=num[i];//未操作的直接保存 
        else 
		{
            R=i;break;//找到被操作的最大位置 
        }
    }
    sort(num+1,num+R+1);//对操作过的数排序便于操作赋值 
    for(int i=R;i>=1;i--)
	{
        if(s2[i]==1&&s1[i]>now)//判断对当前的数的操作的顺序是否大于前一个数如果大于则记录对此数的操作类型
		{
            now=s1[i];//位置 
            nowoper=s2[i];//类型 
        }
        if(nowoper==1)//从小到大，保存继续 
		{
            p[i]=num[R];R--;
        }
        else 
		{
            p[i]=num[F];F++;//从大到小（倒着赋值） 
        }
    }
    for(int i=1;i<=n;++i)
	{
        if(i==n)  printf("%d\n",p[i]);
        else   printf("%d ",p[i]);
    }
    return 0;
}