Problem: 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
&gt;= 1 Y
&lt; 3 N
&lt;= -3 N
&gt; 55 N


Output

17


Input

2
&gt; 100 Y
&lt; -100 Y


Output

Impossible

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
char s[10];
char v;
int a, b, c, n, x;
int main() {
  scanf("%d", &n);
  a = -2000000000;
  b = 2000000000;
  for (; n; --n) {
    scanf(" %s %d %c", s, &x, &v);
    if (s[0] == '<') {
      if (s[1] == '\0') --x;
      if (v == 'N') {
        s[0] = '>';
        ++x;
      }
    } else {
      if (s[1] == '\0') ++x;
      if (v == 'N') {
        s[0] = '<';
        --x;
      }
    }
    if (s[0] == '<' && b > x) b = x;
    if (s[0] == '>' && a < x) a = x;
  }
  if (a <= b)
    printf("%d\n", a);
  else
    printf("Impossible\n");
  return 0;
}