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.

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).

Print the number of unread notifications after each event.

3 4
1 3
1 1
1 2
2 3

1
2
3
2

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

1
2
3
0
1
2

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. 題意：現在給n個應用,三種操作,第一種是一個應用產生一個訊息,第二個是讀完這種應產生的所有的訊息,第三個是讀完前t個訊息;

#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];//要定義在主函式外面，否則會越界 
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)//應用程式b產生一條資訊 
			{
				num[b]++;//應用程式b產生的資訊+1； 
				sum++;//資訊總數+1； 
				ans[++cnt]=b;//第cnt條資訊是由應用程式b產生 
				q[b].push(cnt);//把應用程式b產生的資訊放入佇列b; 
			}
			else if(a==2)//讀完應用程式b產生的所有資訊 
			{
				while(!q[b].empty())
				{
					int k=q[b].front();
					q[b].pop();
					vis[k]=1;//標記第k條資訊已經讀過，防止a=3是重複減去 
				}
				sum-=num[b];//總數減去應用程式b所產生的所有資訊 
				num[b]=0;//清零 
			}
			else//讀完前b條資訊，注意可能讀重複 
			{
				for(int i=now+1;i<=b;i++)
				{
					if(vis[i])//已經讀過 
					continue;
					int temp=ans[i];//判斷該條資訊是由哪個應用程式產生 
					q[temp].pop();
					num[temp]--;
					sum--;
					vis[i]=1;//標記讀過 
				}
			    now=max(b,now);
			}
			printf("%d\n",sum);
		}
	}
	return 0;
}