Codeforces Round #366 (Div. 2) C. Thor （模擬）

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. Iftypei = 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:

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).

本身作的時候沒想到用隊列， 我只想到了從3到2的影響的解決方案，想不出來從2的到3的解決方案。QAQ

本題用隊列作。

對於1，用sum隊列記錄每一個進入的編號 cnt++，同時每一個應用也要開的隊列，把此編號加入到每一個應用的隊列中。

對於2，開一個used數組，若是used[vis[y].front()]=0,那麼此條消息沒看過，更新used值，ans++, pop。

對於3，從總的隊列sum出，直到隊列爲空或者當前編號小於y。

結果cnt-ans.

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <map>
 7 #include <queue>
 8 using namespace std;
 9 const int maxn = 3e5+5;
10 queue<int> sum,vis[maxn];
11 int used[maxn];
12 int main()
13 {
14     int n,q;
15     cin>>n>>q;
16     int ans = 0;
17     int cnt = 0;
18     int x,y;
19     for(int i=1;i<=q;i++)
20     {
21         scanf("%d %d",&x,&y);
22         if(x==1)
23         {
24             cnt++;
25             vis[y].push(cnt);
26             sum.push(cnt);
27         }
28         else if(x==2)
29         {
30             while(!vis[y].empty())
31             {
32                 if(!used[vis[y].front()])
33                 {
34                     used[vis[y].front()] = 1;
35                     ans++;
36                 }
37                 vis[y].pop();
38             }
39         }
40         else
41         {
42             while(!sum.empty()&&sum.front()<=y)
43             {
44 
45                 if(!used[sum.front()])
46                 {
47                     used[sum.front()] = 1;
48                     ans++;
49                 }
50                 sum.pop();
51             }
52         }
53         printf("%d\n",cnt-ans);
54     }
55     return 0;
56 }