C. Books Queries

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

You are given qq queries of three types:

1. L idid — put a book having index idid on the shelf to the left from the leftmost existing book;
2. R idid — put a book having index idid on the shelf to the right from the rightmost existing book;
3. ? idid — 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 idid 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 33 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 idids don't repeat in queries of first two types.

Your problem is to answer all the queries of type 33 in order they appear in the input.

Note that after answering the query of type 33 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 qq (1≤q≤2⋅1051≤q≤2⋅105) — the number of queries.

Then qq lines follow. The ii-th line contains the ii-th query in format as in the problem statement. It is guaranteed that queries are always valid (for query type 33, 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 33 in the input.

In each query the constraint 1≤id≤2⋅1051≤id≤2⋅105 is met.

Output

Print answers to queries of the type 33 in order they appear in the input.

Examples

input

Copy

8
L 1
R 2
R 3
? 2
L 4
? 1
L 5
? 1

output

Copy

1
1
2

input

Copy

10
L 100
R 100000
R 123
L 101
? 123
L 10
R 115
? 100
R 110
? 115

output

Copy

0
2
1

Note

Let's take a look at the first example and let's consider queries:

1. The shelf will look like [1][1];
2. The shelf will look like [1,2][1,2];
3. The shelf will look like [1,2,3][1,2,3];
4. The shelf looks like [1,2,3][1,2,3] so the answer is 11;
5. The shelf will look like [4,1,2,3][4,1,2,3];
6. The shelf looks like [4,1,2,3][4,1,2,3] so the answer is 11;
7. The shelf will look like [5,4,1,2,3][5,4,1,2,3];
8. The shelf looks like [5,4,1,2,3][5,4,1,2,3] so the answer is 22.

Let's take a look at the second example and let's consider queries:

1. The shelf will look like [100][100];
2. The shelf will look like [100,100000][100,100000];
3. The shelf will look like [100,100000,123][100,100000,123];
4. The shelf will look like [101,100,100000,123][101,100,100000,123];
5. The shelf looks like [101,100,100000,123][101,100,100000,123] so the answer is 00;
6. The shelf will look like [10,101,100,100000,123][10,101,100,100000,123];
7. The shelf will look like [10,101,100,100000,123,115][10,101,100,100000,123,115];
8. The shelf looks like [10,101,100,100000,123,115][10,101,100,100000,123,115] so the answer is 22;
9. The shelf will look like [10,101,100,100000,123,115,110][10,101,100,100000,123,115,110];
10. The shelf looks like [10,101,100,100000,123,115,110][10,101,100,100000,123,115,110] so the answer is 11.

題目大意：給你一個書架，讓你按照指令把書放上去，如果輸入L,id，表示把編號為id的書放在書架的最右邊，如果輸入R，id，表示把編號為id的書放在書架的最左邊，如果輸入？id，則求出如果把編號為id的書移動到最右邊或者最左邊需要多少步。

題目大致思路：我們統計一下放在左邊的書有多少個，放在右邊的書有多少個，並且放入當前id的書為於左邊，或者右邊的幾個最後分減一下左邊，右邊的總數（取絕對值），最後取其中的最小值就可以了。

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 2000010;
int pos[MAXN];

int main(){
	int n;
	while(scanf("%d",&n)){
		char op[2];
		int id;
		scanf("%s%d",op,&id);
		pos[id] = 0;
		int CntLeft = 0,CntRight = 0;
		CntLeft--;
		CntRight++;
		for(int i = 2; i <= n; i++){
			scanf("%s%d",op,&id);
			if(op[0] == 'L'){
				pos[id] = CntLeft;
				CntLeft--;
			}
			if(op[0] == 'R'){
				pos[id] = CntRight;
				CntRight++;
			}
			if(op[0] == '?'){
				printf("%d\n",min(abs(pos[id] - CntLeft),abs(pos[id] - CntRight)) - 1);
			}
		}
	}
	return 0;
}