快乐补题-SDNU_ACM_ICPC_2020_Winter_Practice_1st——F

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 = 2 then 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:

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

In the second sample test:
Application 2 generates a notification (there is 1 unread notification).
Application 4 generates a notification (there are 2 unread notifications).
Application 2 generates a notification (there are 3 unread notifications).
Thor reads first three notifications and since there are only three of them so far, there will be no unread notification left.
Application 3 generates a notification (there is 1 unread notification).
Application 3 generates a notification (there are 2 unread notifications).

题目大意：

此题背景是手机app产生未读消息，有n款app，对应三种事件，事件一，x号app产生一条新的未读消息。事件二，雷神读了x号app的所有未读信息。事件三，雷神读了最开始的t条消息，（这些就是按顺序产生的app消息，不管读没读）。每次事件后，都要输出当前的未读消息数。

思路分析：

开数个队列，分别记录总消息以及各个app的消息，然后一顿操作bulabula就阔以了

AC代码：

#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define INF 0x3f3f3f3f3f3f3f 
#define MOD 1000000007
#define MAXN 300005
using namespace std;
inline LL read()
{
    LL x=0,w=1;
    char ch=0;
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')
            w=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return w*x;
}
queue<int> a,vis[MAXN];
int n,q,ans,cnt,x,y;
int readd[MAXN];
int main()
{
    n=read();
    q=read();
    for(int i=1;i<=q;i++)
    {
    	x=read();
        y=read();
        if(x==1)
        {
            cnt++;
            vis[y].push(cnt);
            a.push(cnt);
        }
       	if(x==2)
    	{
            while(!vis[y].empty())
            {
               if(!readd[vis[y].front()])
                {
                    readd[vis[y].front()]=1;
                    ans++;
                }
                vis[y].pop();
            }
        }
        if(x==3)
        {
            while(!a.empty()&&a.front()<=y)
            {
                if(!readd[a.front()])
                {
                    readd[a.front()]=1;
                    ans++;
                }
               a.pop();
           }
        }
        printf("%d\n",cnt-ans);
    }
    return 0;
}