Codeforces Round #366 (Div. 2) C 模拟queue

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).

题意：n个app q个事件 三种操作 

        1  x  应用x增加一个未读消息

        2  x  应用x的消息全部被阅读

        3  t   前t个 1操作的事件全部被阅读

q个操作 每次输出当前剩余的未读信息的数量

题解：队列模拟

 un[i]表示应用i总的信息数量

com[i]表示应用i的已读的信息数量

/******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^    ^ ^
 O      O
******************************/
//#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<algorithm>
#include<queue>
using namespace std;
#define ll __int64
#define esp 1e-10
const int N=3e5+10,M=1e6+10,mod=1e9+7,inf=1e9+10;
int a[N];
int com[N];
int un[N];
int th[N];
queue<int>q;
int main()
{
    int x,y,z,i,t;
    scanf("%d%d",&x,&y);
    int ans=0,maxx=0;
    for(i=0;i<y;i++)
    {
        int u;
        scanf("%d%d",&u,&a[i]);
        if(u==1)
        un[a[i]]++,ans++,q.push(a[i]);
        else if(u==2)
        {
            ans-=un[a[i]]-com[a[i]];
            com[a[i]]=un[a[i]];
        }
        else
        {
            if(a[i]>=maxx)//只需要出队入队一次
            {
                int T=a[i]-maxx;
                while(!q.empty()&&T>0)
                {
                    T--;
                    int v=q.front();
                    q.pop();
                    th[v]++;
                    if(th[v]>com[v])//执行3操作的阅读量大于已经阅读的量
                    {
                        ans-=th[v]-com[v];
                        com[v]=th[v];
                    }
                }
                maxx=a[i];
            }
        }
        printf("%d
",ans);
    }
    return 0;
}