题干：

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:

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).

题目大意：

第一行两个整数n，m，代表n个软件，m次操作。(1 ≤ n, q ≤ 300 000) 

接下来m行：
1 x :软件 x 产生了一条未读信息。
2 x: 软件 x 的所有信息都被读了。
3  t: 前 t 条产生的信息都被读了。注意是前t条，不是未读的前t条，已读的也算在t条中。

解题报告：

  本来想用个双向链表来啊维护出现过的所有信息，但是发现它可以支持快速的删除一个元素，但是对于查找前t条信息比较鸡肋，甚至还要拿个并查集去搞，比较麻烦，于是换了各项性能都还说的过去的set，直接二分找前t条信息就可以了，这样最后的答案就是set的大小。

 注意不能让标号直接等于i，因为第i个操作不一定是产生一天新的信息。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 5e5 + 5;
int n,m,last,id;
set<int> ss[MAX],q;//存链表，因为值直接就是索引就可以了，不需要单独弄链表了int main()
{cin>>n>>m;for(int op,x,i = 1; i<=m; i++) {scanf("%d%d",&op,&x);if(op == 1) {id++;ss[x].insert(id);q.insert(id);}if(op == 2) {for(auto idx : ss[x]) {if(q.find(idx) != q.end())q.erase(idx);}ss[x].clear();}if(op == 3) {auto upp = q.upper_bound(x),down = q.begin();q.erase(down,upp);last = max(last,x);}printf("%d\n",q.size());}return 0 ;
}