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 long n, m = 0, k = 0, j = 0;
bool test = false, test1 = false;
;
long long const N = 1e5 + 7;
long long const oo = 1e5 + 7;
vector<long long> v, v2;
map<pair<long long, long long>, long long> my;
vector<long long> v1[oo];
long long z[oo];
void dfs(long long s) {
  if (z[s]) {
    return;
  }
  z[s] = 1;
  if (v1[s].size() == 1 && s != 0 && k != 0) {
    m++;
    k = 0;
    v.push_back(s + 1);
  }
  for (long long u = 0; u < v1[s].size(); u++) {
    if (my[make_pair(s, v1[s][u])] == 2) {
      k++;
      my[make_pair(s, v1[s][u])] = 1;
      my[make_pair(v1[s][u], s)] = 1;
    }
    dfs(v1[s][u]);
  }
}
vector<pair<long long, long long> > vv;
map<char, long long> my1;
priority_queue<long long> q;
long long t[N];
long long dx[4] = {0, 1, 0, -1};
long long dy[4] = {1, 0, -1, 0};
bool in_da_prison(long long x, long long y) {
  return (x < 9 && x >= 0 && y < 9 && y >= 0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long i = 0, j = 0, a = 0, b = 0, c = 0, d = 0, k = 0;
  string s1, s2, s3;
  cin >> n;
  a = -2 * pow(10, 9);
  b = 2 * pow(10, 9);
  char ch;
  for (i = 0; i < n; i++) {
    cin >> s1 >> k >> ch;
    if (s1 == ">=") {
      if (ch == 'Y') {
        a = max(k, a);
      } else {
        b = min(b, k - 1);
      }
    } else if (s1 == "<=") {
      if (ch == 'Y') {
        b = min(k, b);
      } else {
        a = max(k + 1, a);
      }
    } else if (s1 == "<") {
      if (ch == 'Y') {
        b = min(k - 1, b);
      } else {
        a = max(k, a);
      }
    } else if (s1 == ">") {
      if (ch == 'Y') {
        a = max(k + 1, a);
      } else {
        b = min(k, b);
      }
    }
  }
  if (b < a) {
    cout << "Impossible";
    return 0;
  }
  cout << b;
  return 0;
}