贪心 C

Polycarp has recently got himself a new job. He now earns so much that his old wallet can't even store all the money he has.

Berland bills somehow come in lots of different sizes. However, all of them are shaped as rectangles (possibly squares). All wallets are also produced in form of rectangles (possibly squares).

A bill x×y fits into some wallet h×w if either x≤h and y≤w or y≤h and x≤w . Bills can overlap with each other in a wallet and an infinite amount of bills can fit into a wallet. That implies that all the bills Polycarp currently have fit into a wallet if every single one of them fits into it independently of the others.

Now you are asked to perform the queries of two types:

1. + x y — Polycarp earns a bill of size x×y ;
2. ? h w — Polycarp wants to check if all the bills he has earned to this moment fit into a wallet of size h×w .

It is guaranteed that there is at least one query of type 1 before the first query of type 2 and that there is at least one query of type 2 in the input data.

For each query of type 2 print "YES" if all the bills he has earned to this moment fit into a wallet of given size. Print "NO" otherwise.

Input

The first line contains a single integer n (2≤n≤5⋅105 ) — the number of queries.

Each of the next n lines contains a query of one of these two types:

1. + x y (1≤x,y≤109 ) — Polycarp earns a bill of size x×y ;
2. ? h w (1≤h,w≤109 ) — Polycarp wants to check if all the bills he has earned to this moment fit into a wallet of size h×w .

It is guaranteed that there is at least one query of type 1 before the first query of type 2 and that there is at least one query of type 2 in the input data.

Output

For each query of type 2 print "YES" if all the bills he has earned to this moment fit into a wallet of given size. Print "NO" otherwise.

Example

Input

9
+ 3 2
+ 2 3
? 1 20
? 3 3
? 2 3
+ 1 5
? 10 10
? 1 5
+ 1 1

Output

NO
YES
YES
YES
NO

Note

The queries of type 2 of the example:

1. Neither bill fits;
2. Both bills fit (just checking that you got that bills can overlap);
3. Both bills fit (both bills are actually the same);
4. All bills fit (too much of free space in a wallet is not a problem);
5. Only bill 1×5 fit (all the others don't, thus it's "NO").

题目大意：

+ 代表给你的money，再给你两个数字，代表长和宽，？代表钱包，问你在出现？前面所有的Money是不是可以装进钱包，可以输出YES，否则输出NO

思路：先对输入数字进行处理，a,b, 输入之后，进行处理，a<b.

出现？，进行查找，很容易可以发现，这里的a一定要比前面所有的a中最大的一个还要大，这里的b同理

否则就装不下所有的。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn=2e5+10;

int main()
{
    int n,maxx=0,maxy=0;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        int a,b;
        char c[10];
        scanf("%s",c);
        scanf("%d%d",&a,&b);
        if(a>b) swap(a,b);
        if(c[0]=='+')
        {
            maxx=max(maxx,a);
            maxy=max(maxy,b);
        }
        if(c[0]=='?')
        {
            if(a>=maxx&&b>=maxy) printf("YES
");
            else printf("NO
");
        }
    }
    return 0;
}