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;
long double dis(double x, double y, double x1, double y1) {
  return sqrt((x - x1) * (x - x1) + (y - y1) * (y - y1));
}
double PI = 3.14159265;
double e = 2.718281828;
void EnGz() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
}
long long GCD_(long long a, long long b) { return b ? GCD_(b, a % b) : a; }
long long LCM_(long long a, long long b) { return (a * b) / GCD_(a, b); }
void print(vector<long long> v) {
  for (int i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << endl;
}
int main() {
  EnGz();
  int mn = -2e9, mx = 2e9, n;
  cin >> n;
  string s;
  char ch;
  int q;
  for (int i = 0; i < n; i++) {
    cin >> s >> q >> ch;
    if (s == ">=") {
      if (ch == 'Y')
        mn = max(q, mn);
      else
        mx = min(q - 1, mx);
    } else if (s == "<=") {
      if (ch == 'Y')
        mx = min(mx, q);
      else
        mn = max(mn, q + 1);
    } else if (s == ">") {
      if (ch == 'Y')
        mn = max(mn, q + 1);
      else
        mx = min(q, mx);
    } else {
      if (ch == 'Y')
        mx = min(mx, q - 1);
      else
        mn = max(q, mn);
    }
  }
  if (mn <= mx)
    cout << mn;
  else
    cout << "Impossible";
  return 0;
}