Codeforces #366(div 2)



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 type_i — type of the i-th event. If type_i = 1 or type_i = 2 then it is followed by an integer x_i. Otherwise it is followed by an integer t_i (1 ≤ type_i ≤ 3, 1 ≤ x_i ≤ n, 1 ≤ t_i ≤ 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).

题面意思还是挺容易理解的，就是每个app会发通知，有三种操作：

     1    某个app发通知

     2    读某个app的全部通知

     3    读前n条通知

这道题主要的重点在于第二种和第三种的去重操作和第三种自己本身的去重操作。

针对一个app的通知，在读过之后应该维护一个位置值，相当于将读过的清空处理。在第三种情况中，如果新的通知的位置大于这个位置，那么就可以进行阅读。最开始WA的原因是没有仔细思考如何进行维护位置。

下面是AC代码和WA代码：

AC代码：

#include<iostream>

using namespace std;
int a[300010];
int pos[300010];
int lr[300010];
int times[300010];
int beg;
int main(){
#ifdef LOCAL
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
#endif
	int n,q;
	cin>>n>>q;
	int temp=0;
	int ans=0;
	beg=0;
	for(int i=1;i<=q;i++){
		int x,y;
		cin>>x>>y;
		if(x==1){
			a[temp]=y;
			pos[temp++]=i;
			times[y]++;
			ans++;
		}
		else if(x==2){
			ans-=times[y];
			times[y]=0;
			lr[y]=i;
		}
		else if(x==3){
			for(int j=beg;j<y;j++){
				if(pos[j]>lr[a[j]]){
					times[a[j]]--;
					ans--;
				}
			}
			beg=max(beg,y);
		}
		cout<<ans<<endl;
	}
	return 0;


}

WA代码：

#include<iostream>
using namespace std;
int a[300010];
int b[300010];
int main(){
#ifdef LOCAL
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
#endif
	int n,q;
	cin>>n>>q;
	int temp=0;
	int ans=0;
	for(int i=0;i<q;i++){
		int x,y;
		cin>>x>>y;
		if(x==1){
			a[temp++]=y;
			b[y]++;
			ans++;
		}
		else if(x==2){
			ans-=b[y];
			b[y]=0;
		}
		else if(x==3){
			for(int i=0;i<y;i++){
				b[a[i]]--;
				ans--;
			}
		}
		cout<<ans<<endl;
	}
	return 0;
}