CodeForces Round #515 Div.3 C. Books Queries

http://codeforces.com/contest/1066/problem/C

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

You are given $\mathrm{qq queries\; of\; three\; types:}$

1. L $\mathrm{idid —\; put\; a\; book\; having\; index idid on\; the\; shelf\; to\; the\; left\; from\; the\; leftmost\; existing\; book;}$
2. R $\mathrm{idid —\; put\; a\; book\; having\; index idid on\; the\; shelf\; to\; the\; right\; from\; the\; rightmost\; existing\; book;}$
3. ? $\mathrm{idid —\; calculate\; the\; minimum\; number\; of\; books\; you\; need\; to\; pop\; from\; the\; left\; or\; from\; the\; right\; in\; such\; a\; way\; that\; the\; book\; with\; index idid will\; be\; leftmost\; or\; rightmost.}$

You can assume that the first book you will put can have any position (it does not matter) and queries of type $\mathrm{33 are\; always\; valid\; (it\; is\; guaranteed\; that\; the\; book\; in\; each\; such\; query\; is\; already\; placed).\; You\; can\; also\; assume\; that\; you\; don\text{'}t\; put\; the\; same\; book\; on\; the\; shelf\; twice,\; so idids\; don\text{'}t\; repeat\; in\; queries\; of\; first\; two\; types.}$

Your problem is to answer all the queries of type $\mathrm{33 in\; order\; they\; appear\; in\; the\; input.}$

Note that after answering the query of type $\mathrm{33 all\; the\; books\; remain\; on\; the\; shelf\; and\; the\; relative\; order\; of\; books\; does\; not\; change.}$

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 $\mathrm{qq (1\le q\le 2\cdot 1051\le q\le 2\cdot 105)\; —\; the\; number\; of\; queries.}$

Then $\mathrm{qq lines\; follow.\; The ii-th\; line\; contains\; the ii-th\; query\; in\; format\; as\; in\; the\; problem\; statement.\; It\; is\; guaranteed\; that\; queries\; are\; always\; valid\; (for\; query\; type 33,\; it\; is\; guaranteed\; that\; the\; book\; in\; each\; such\; query\; is\; already\; placed,\; and\; for\; other\; types,\; it\; is\; guaranteed\; that\; the\; book\; was\; not\; placed\; before).}$

It is guaranteed that there is at least one query of type $\mathrm{33 in\; the\; input.}$

In each query the constraint $\mathrm{1\le id\le 2\cdot 1051\le id\le 2\cdot 105 is\; met.}$

Output

Print answers to queries of the type $\mathrm{33 in\; order\; they\; appear\; in\; the\; input.}$

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 $[10,101,100,100000,123,115,110][10,101,100,100000,123,115,110] so\; the\; answer\; is 11.$

代码：

#include <iostream>
#include <cstring>
#include <stdio.h>
using namespace std;

int a[200005];

int main() {
    int q;
    while (~scanf("%d", &q)) {
        char op[2];
        memset(a, 0, sizeof(a));
        int x, l = 100000, r = 100000;
        for (int i = 0; i < q; i ++) {
            scanf("%s%d", op, &x);
            if (strcmp(op, "L") == 0) {
                a[x] = l --;
                if (i == 0) r ++;
            } else if (strcmp(op, "R") == 0) {
                a[x] = r ++;
                if (i == 0) l --;
            }
            else
                printf("%d
", min(a[x] - l, r - a[x]) - 1);

        }
    }
    return 0;
}