You have got a shelf and want to put some books on it.

You are given q queries of three types:

L
id — put a book having index id on the shelf to the left from the leftmost existing book;
R id — put a book having index id on the shelf to the right from the rightmost existing book;
? id — calculate the minimum number of books you need to pop from the left or from the right in such a way that the book with index id will be leftmost or rightmost.
You can assume that the first book you will put can have any position (it does not matter) and queries of type 3 are always valid (it is guaranteed that the book in each such query is already placed). You can also assume that you don’t put the same book on the shelf twice, so ids don’t repeat in queries of first two types.

Your problem is to answer all the queries of type 3 in order they appear in the input.

Note that after answering the query of type 3 all the books remain on the shelf and the relative order of books does not change.

If you are Python programmer, consider using PyPy instead of Python when you submit your code.

Input
The first line of the input contains one integer （1≤q≤2⋅105) — the number of queries.

Then q lines follow. The i-th line contains the i-th query in format as in the problem statement. It is guaranteed that queries are always valid (for query type 3, it is guaranteed that the book in each such query is already placed, and for other types, it is guaranteed that the book was not placed before).

It is guaranteed that there is at least one query of type 3 in the input.

In each query the constraint 1≤id≤2⋅105 is met.

Output
Print answers to queries of the type。3 in order they appear in the input.

Examples
inputCopy
8
L 1
R 2
R 3
? 2
L 4
? 1
L 5
? 1
outputCopy
1
1
2
inputCopy
10
L 100
R 100000
R 123
L 101
? 123
L 10
R 115
? 100
R 110
? 115
outputCopy
0
2
1
Note
Let’s take a look at the first example and let’s consider queries:

The shelf will look like [1];
The shelf will look like [1,2];
The shelf will look like [1,2,3];
The shelf looks like [1,2,3] so the answer is 1;
The shelf will look like [4,1,2,3];
The shelf looks like [4,1,2,3] so the answer is 1;
The shelf will look like [5,4,1,2,3];
The shelf looks like [5,4,1,2,3] so the answer is 2.
Let’s take a look at the second example and let’s consider queries:

The shelf will look like [100];
The shelf will look like [100,100000];
The shelf will look like [100,100000,123];
The shelf will look like [101,100,100000,123];
The shelf looks like [101,100,100000,123] so the answer is 0;
The shelf will look like [10,101,100,100000,123];
The shelf will look is[10,101,100,100000,123,115];
The shelf looks like [10,101,100,100000,123,115] so the answer is 2;
The shelf will look like [10,101,100,100000,123,115,110];
The shelf looks like [10,101,100,100000,123,115,110] so the answer is 1.

雖然是c題了，但是感覺還是沒有那麼難，比第二個簡單。。
就是按著左或者右來放書，？的時候就查尋從左或者從右最小的位置。用兩個數標記左右的位置就行了。
程式碼如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxx=2e5+10;
int a[maxx];

int main()
{
	int q;
	char s;
	int p;
	while(cin>>q)
	{
		int l=1;
		int r=0;
		for(int i=0;i<q;i++)
		{
			cin>>s>>p;
			if(s=='L') a[p]=--l;
			else if(s=='R') a[p]=++r;
			else cout<<min(a[p]-l,r-a[p])<<endl;
		}
	}
}

努力加油a啊，(^o)/~