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;
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
char op[3], fuck_math_h[2];
int main() {
  int N;
  scanf("%d", &N);
  int L = -2000000000, R = 2000000000;
  for (int(i) = 0; (i) < (int)(N); ++(i)) {
    int x;
    scanf("%s%d%s", op, &x, fuck_math_h);
    if (op[0] == '>' && op[1] == 0) {
      if (*fuck_math_h == 'Y')
        amax(L, x + 1);
      else
        amin(R, x);
    } else if (op[0] == '<' && op[1] == 0) {
      if (*fuck_math_h == 'Y')
        amin(R, x - 1);
      else
        amax(L, x);
    } else if (op[0] == '>' && op[1] == '=') {
      if (*fuck_math_h == 'Y')
        amax(L, x);
      else
        amin(R, x - 1);
    } else if (op[0] == '<' && op[1] == '=') {
      if (*fuck_math_h == 'Y')
        amin(R, x);
      else
        amax(L, x + 1);
    }
  }
  if (L > R)
    puts("Impossible");
  else
    printf("%d\n", L);
  return 0;
}