Codeforces Round #241 (Div. 2)->A. Guess a number!

A. Guess a number!

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show.

The rules are simple. The host thinks of an integer y and the participants guess it by asking questions to the host. There are four types of acceptable questions:

• Is it true that y is strictly larger than number x?
• Is it true that y is strictly smaller than number x?
• Is it true that y is larger than or equal to number x?
• Is it true that y is smaller than or equal to number x?

On each question the host answers truthfully, "yes" or "no".

Given the sequence of questions and answers, find any integer value of y that meets the criteria of all answers. If there isn't such value, print "Impossible".

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 10000) — the number of questions (and answers). Next n lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is:

• ">" (for the first type queries),
• "<" (for the second type queries),
• ">=" (for the third type queries),
• "<=" (for the fourth type queries).

All values of x are integer and meet the inequation  - 109 ≤ x ≤ 109. The answer is an English letter "Y" (for "yes") or "N" (for "no").

Consequtive elements in lines are separated by a single space.

Output

Print any of such integers y, that the answers to all the queries are correct. The printed number y must meet the inequation - 2·109 ≤ y ≤ 2·109. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes).

Examples

input

4
>= 1 Y
< 3 N
<= -3 N
> 55 N

output

17

input

2
> 100 Y
< -100 Y

output

Impossible

题意理解：猜数字，输出猜出的数字中任意一个，如果不符合则输出“Impossible”。

 #include<bits/stdc++.h>
 using namespace std;
 const int INF=;
 int main()
 {
     int n,x;
     cin>>n;
     int high=INF,low=-INF;//取临界值
     for(int i=; i<n; i++)
     {
         char s[]= {},t[];
         scanf("%s%d%s",s,&x,t);
         if(t[]=='N')
         {
             s[]^='=';
             s[]^='<'^'>';
         }//如果是N的话则把它取反
         if(s==string(">")) low=max(low,x+);//前面用char后面得强制转换一下
         if(s==string("<"))  high=min(high,x-);
         if(s==string(">=")) low=max(low,x);
         if(s==string("<=")) high=min(high,x);
     }//大取大小取小
     if(low<=high) printf("%d\n",low);
     else printf("Impossible\n");
     return ;
 }