B - Guess a number!

Problem description

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 signis:

• ">" (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).

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Examples

Input

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

Output

17

Input

2
> 100 Y
< -100 Y

Output

Impossible
解题思路：题意比较简单，就是找出符合要求的区间，然后输出其中任意一个值。注意有可能只有左区间或者只有右区间，若是这样，各做一个标记即可。
AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int INF = 2e9;
 4 int main(){
 5     int n,x,left=-INF,right=INF;char ch;string eq;bool flag1=false,flag2=false;//[left,right]
 6     cin>>n;
 7     while(n--){
 8         getchar();
 9         cin>>eq>>x>>ch;
10         if(ch=='N'){//修改其值
11             if(eq==">=")eq="<";
12             else if(eq=="<=")eq=">";
13             else if(eq==">")eq="<=";
14             else eq=">=";
15         }
16         if(eq=="<"){right=min(right,x-1);flag2=true;}//右区间取最小
17         else if(eq=="<="){right=min(right,x);flag2=true;}
18         else if(eq==">"){left=max(left,x+1);flag1=true;}//左区间取最大
19         else {left=max(left,x);flag1=true;}
20     }
21     if(left>right)cout<<"Impossible"<<endl;
22     else if(flag1)cout<<left<<endl;//可能只有左区间
23     else if(flag2)cout<<right<<endl;//可能只有右区间
24     return 0;
25 }