Codeforces Round #366 (Div. 2) C 模擬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).

題意：n個app q個事件 三種操做 

        1  x  應用x增長一個未讀消息

        2  x  應用x的消息所有被閱讀

        3  t   前t個 1操做的事件所有被閱讀

q個操做 每次輸出當前剩餘的未讀信息的數量

題解：隊列模擬

 un[i]表示應用i總的信息數量

com[i]表示應用i的已讀的信息數量

 1 /******************************
 2 code by drizzle
 3 blog: www.cnblogs.com/hsd-/
 4 ^ ^    ^ ^
 5  O      O
 6 ******************************/
 7 //#include<bits/stdc++.h>
 8 #include<iostream>
 9 #include<cstring>
10 #include<cstdio>
11 #include<map>
12 #include<algorithm>
13 #include<queue>
14 using namespace std;
15 #define ll __int64
16 #define esp 1e-10
17 const int N=3e5+10,M=1e6+10,mod=1e9+7,inf=1e9+10;
18 int a[N];
19 int com[N];
20 int un[N];
21 int th[N];
22 queue<int>q;
23 int main()
24 {
25     int x,y,z,i,t;
26     scanf("%d%d",&x,&y);
27     int ans=0,maxx=0;
28     for(i=0;i<y;i++)
29     {
30         int u;
31         scanf("%d%d",&u,&a[i]);
32         if(u==1)
33         un[a[i]]++,ans++,q.push(a[i]);
34         else if(u==2)
35         {
36             ans-=un[a[i]]-com[a[i]];
37             com[a[i]]=un[a[i]];
38         }
39         else
40         {
41             if(a[i]>=maxx)//只須要出隊入隊一次
42             {
43                 int T=a[i]-maxx;
44                 while(!q.empty()&&T>0)
45                 {
46                     T--;
47                     int v=q.front();
48                     q.pop();
49                     th[v]++;
50                     if(th[v]>com[v])//執行3操做的閱讀量大於已經閱讀的量
51                     {
52                         ans-=th[v]-com[v];
53                         com[v]=th[v];
54                     }
55                 }
56                 maxx=a[i];
57             }
58         }
59         printf("%d\n",ans);
60     }
61     return 0;
62 }