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).

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。如果不满足构成区间的条件，则输出"Impossible"。
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;//初始值区间满足-2*10^9<=[left,right]<=2*10^9
 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);//取右端点的最小值
17         else if(eq=="<=")right=min(right,x);
18         else if(eq==">")left=max(left,x+1);//取左端点的最大值
19         else left=max(left,x);
20     }
21     if(left>right)cout<<"Impossible"<<endl;//不满足条件
22     else cout<<left<<endl;//直接输出左端点值（任意值）
23     return 0;
24 }











            

            
posted @ 
2018-06-05 22:50 
霜雪千年 
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