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 }