Social Network (hard version) CodeForces - 1234B2（离散化+双端队列）-白红宇

Social Network (hard version) CodeForces - 1234B2（离散化+双端队列）

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发布时间：2019-05-25

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The only difference between easy and hard versions are constraints on nn and kk.

You are messaging in one of the popular social networks via your smartphone. Your smartphone can show at most kk most recent conversations with your friends. Initially, the screen is empty (i.e. the number of displayed conversations equals 00).

Each conversation is between you and some of your friends. There is at most one conversation with any of your friends. So each conversation is uniquely defined by your friend.

You (suddenly!) have the ability to see the future. You know that during the day you will receive nn messages, the ii-th message will be received from the friend with ID idiidi (1≤idi≤1091≤idi≤109).

If you receive a message from idiidi in the conversation which is currently displayed on the smartphone then nothing happens: the conversations of the screen do not change and do not change their order, you read the message and continue waiting for new messages.

Otherwise (i.e. if there is no conversation with idiidi on the screen):

Firstly, if the number of conversations displayed on the screen is kk, the last conversation (which has the position kk) is removed from the screen.

Now the number of conversations on the screen is guaranteed to be less than kk and the conversation with the friend idiidi is not displayed on the screen.

The conversation with the friend idiidi appears on the first (the topmost) position on the screen and all the other displayed conversations are shifted one position down.

Your task is to find the list of conversations (in the order they are displayed on the screen) after processing all nn messages.

Input

The first line of the input contains two integers nn and kk (1≤n,k≤2⋅105)1≤n,k≤2⋅105) — the number of messages and the number of conversations your smartphone can show.

The second line of the input contains nn integers id1,id2,…,idnid1,id2,…,idn (1≤idi≤1091≤idi≤109), where idiidi is the ID of the friend which sends you the ii-th message.

Output

In the first line of the output print one integer mm (1≤m≤min(n,k)1≤m≤min(n,k)) — the number of conversations shown after receiving all nn messages.

In the second line print mm integers ids1,ids2,…,idsmids1,ids2,…,idsm, where idsiidsi should be equal to the ID of the friend corresponding to the conversation displayed on the position ii after receiving all nn messages.

Examples

Input

7 2

1 2 3 2 1 3 2

Output

2 1

Input

10 4

2 3 3 1 1 2 1 2 3 3

Output

1 3 2

Note

In the first example the list of conversations will change in the following way (in order from the first to last message):

[][];

[1][1];

[2,1][2,1];

[3,2][3,2];

[1,3][1,3];

[2,1][2,1].

In the second example the list of conversations will change in the following way:

[][];

[2][2];

[3,2][3,2];

[1,3,2][1,3,2];

and then the list will not change till the end.

这个数据量就有点大了。再用vector就超时了。（vector比deque慢的不是一点半点）

我们离散化一下，就可以用数组去表示每个数字的个数。然后用双端队列进行增减操作。

代码如下：

#include
   
    #define ll long longusing namespace std;const int maxx=2e5+100;int a[maxx];int b[maxx];int vis[maxx];int n,m;int main(){
   	scanf("%d%d",&n,&m);	for(int i=1;i<=n;i++) scanf("%d",&a[i]),b[i]=a[i];	sort(b+1,b+1+n);	int len=unique(b+1,b+1+n)-b-1;	for(int i=1;i<=n;i++) a[i]=lower_bound(b+1,b+1+len,a[i])-b;	deque
    
      h;	for(int i=1;i<=n;i++)	{
   		if(vis[a[i]]) continue;		else		{
   			if(h.size()

努力加油a啊，(^o)/~

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