CodeForces - 416A (判断大于小于等于 模拟题)

Guess a number!

Time Limit: 1000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit Status

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 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).

Sample Input

Input

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

Output

17

Input

2
> 100 Y
< -100 Y

Output

Impossible

Source
Codeforces Round #241 (Div. 2)

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
int f[100005];

int main()
{
int n,min0=-1000001000,max0=1000010000;
cin>>n;
for(int i=0;i<n;i++)
{
char x[3],an;int num;
cin>>x>>num>>an;
if(an=='Y'&&x[0]=='>')
{
if(x[1]=='=')
min0=max(min0,num);
else min0=max(min0,num+1);
}
else if(an=='N'&&x[0]=='>')
{
if(x[1]=='=')
max0=min(max0,num-1);
else max0=min(max0,num);
}
else if(an=='Y'&&x[0]=='<')
{
if(x[1]=='=')
max0=min(max0,num);
else max0=min(max0,num-1);
}
else if(an=='N'&&x[0]=='<')
{
if(x[1]=='=')
min0=max(min0,num+1);
else min0=max(min0,num);
}
// cout<<max0<<' '<<min0<<endl;
}
if(max0>=min0)
cout<<min0<<endl;
else cout<<"Impossible"<<endl;
return 0;
}