CodeForces 705C Thor （模拟）

C. Thor

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Thor is getting used to the Earth. As a gift Loki gave him a smartphone. There are n applications on this phone. Thor is fascinated by this phone. He has only one minor issue: he can't count the number of unread notifications generated by those applications (maybe Loki put a curse on it so he can't).

q events are about to happen (in chronological order). They are of three types:

1. Application x generates a notification (this new notification is unread).
2. Thor reads all notifications generated so far by application x (he may re-read some notifications).
3. Thor reads the first t notifications generated by phone applications (notifications generated in first t events of the first type). It's guaranteed that there were at least t events of the first type before this event. Please note that he doesn't read first t unread notifications, he just reads the very first t notifications generated on his phone and he may re-read some of them in this operation.

Please help Thor and tell him the number of unread notifications after each event. You may assume that initially there are no notifications in the phone.

Input

The first line of input contains two integers n and q (1 ≤ n, q ≤ 300 000) — the number of applications and the number of events to happen.

The next q lines contain the events. The i-th of these lines starts with an integer typei — type of the i-th event. If typei = 1 or typei = 2then it is followed by an integer xi. Otherwise it is followed by an integer ti (1 ≤ typei ≤ 3, 1 ≤ xi ≤ n, 1 ≤ ti ≤ q).

Output

Print the number of unread notifications after each event.

Examples

input

3 4
1 3
1 1
1 2
2 3

output

1
2
3
2

input

4 6
1 2
1 4
1 2
3 3
1 3
1 3

output

1
2
3
0
1
2

Note

In the first sample:

1. Application 3 generates a notification (there is 1 unread notification).
2. Application 1 generates a notification (there are 2 unread notifications).
3. Application 2 generates a notification (there are 3 unread notifications).
4. Thor reads the notification generated by application 3, there are 2 unread notifications left.

In the second sample test:

1. Application 2 generates a notification (there is 1 unread notification).
2. Application 4 generates a notification (there are 2 unread notifications).
3. Application 2 generates a notification (there are 3 unread notifications).
4. Thor reads first three notifications and since there are only three of them so far, there will be no unread notification left.
5. Application 3 generates a notification (there is 1 unread notification).
6. Application 3 generates a notification (there are 2 unread notifications).
7. 题意：现在给n个应用,三种操作,第一种是一个应用产生一个消息,第二个是读完这种应产生的所有的消息,第三个是读完前t个消息;
8. 
   

用队列模拟过程。注意可能读重复。

#include<stdio.h>
#include<string.h>
#include<queue>
#include<vector>
#include<algorithm>
#define MAX 300000+10
using namespace std;
int num[MAX];
int ans[MAX];
int vis[MAX]; 
queue<int>q[MAX];//要定义在主函数外面，否则会越界 
int n,Q,a,b;
int main()
{
	scanf("%d%d",&n,&Q);
	{
		int cnt=0;
		int sum=0;
		int now=0;
		memset(vis,0,sizeof(vis));
		memset(ans,0,sizeof(ans));
		memset(num,0,sizeof(num));
		for(int i=1;i<=Q;i++)
		{
			scanf("%d%d",&a,&b);
			if(a==1)//应用程序b产生一条信息 
			{
				num[b]++;//应用程序b产生的信息+1； 
				sum++;//信息总数+1； 
				ans[++cnt]=b;//第cnt条信息是由应用程序b产生 
				q[b].push(cnt);//把应用程序b产生的信息放入队列b; 
			}
			else if(a==2)//读完应用程序b产生的所有信息 
			{
				while(!q[b].empty())
				{
					int k=q[b].front();
					q[b].pop();
					vis[k]=1;//标记第k条信息已经读过，防止a=3是重复减去 
				}
				sum-=num[b];//总数减去应用程序b所产生的所有信息 
				num[b]=0;//清零 
			}
			else//读完前b条信息，注意可能读重复 
			{
				for(int i=now+1;i<=b;i++)
				{
					if(vis[i])//已经读过 
					continue;
					int temp=ans[i];//判断该条信息是由哪个应用程序产生 
					q[temp].pop();
					num[temp]--;
					sum--;
					vis[i]=1;//标记读过 
				}
			    now=max(b,now);
			}
			printf("%d\n",sum);
		}
	}
	return 0;
}