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;
int N, P, L, R, F, i, j;
string S, T[4] = {">", ">=", "<", "<="};
map<string, int> M;
char C;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  R = 1e9 + 1;
  L = -R;
  for (i = 0; i < 4; i++) {
    M[T[i]] = i;
  }
  cin >> N;
  for (i = 0; i < N; i++) {
    cin >> S >> P >> C;
    F = M[S];
    if (C == 'N') {
      F = 3 - F;
    }
    if (F < 2) {
      L = max(L, P + 1 - F);
    } else {
      R = min(R, P - 3 + F);
    }
  }
  if (L > R) {
    cout << "Impossible\n";
  } else {
    cout << (L + R) / 2 << '\n';
  }
}