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#include <iostream>
#include <fstream>
#include <string>
std::ifstream f("bool.in");
std::ofstream g("bool.out");
int i,map[256],t;
char var;
std::string s;
void parseString(std::string& s){
std::string sol;
for(int i = 0;i < s.size();++i){
if(s[i] == ' ')
continue;
if(s[i] == '(' || s[i] == ')'){
sol += s[i];
continue;
}
if(s[i] == 'T' && s[i + 1] == 'R'){
sol += '1';
i += 3;
continue;
}
if(s[i] == 'F' && s[i + 1] == 'A'){
sol += '0';
i += 4;
continue;
}
if(s[i] == 'A' && s[i + 1] == 'N'){
sol += '&';
i += 2;
continue;
}
if(s[i] == 'N' && s[i + 1] == 'O'){
sol += "1^";
i += 2;
continue;
}
if(s[i] == 'O' && s[i + 1] == 'R'){
sol += '|';
i++;
continue;
}
sol += s[i];
}
s = sol;
}
bool expresieOr();
bool expresieAnd();
bool expresieXor();
bool factor();
bool expresieOr(){
bool r = expresieAnd();
while(s[i] == '|'){
i++;
r = (r | expresieAnd());
}
return r;
}
bool expresieAnd(){
bool r = expresieXor();
while(s[i] == '&'){
i++;
r = (r & expresieXor());
}
return r;
}
bool expresieXor(){
bool r = factor();
while(s[i] == '^'){
i++;
r = (r ^ factor());
}
return r;
}
bool factor(){
bool r;
if(s[i] == '('){
i++;
r = expresieOr();
i++;
}else{
if(s[i] == '1'){
i++;
return 1;
}
if(s[i] == '0'){
i++;
return 0;
}
return map[s[i++]];
}
return r;
}
int main(){
std::getline(f,s);
parseString(s);
f >> t;
while(t--){
f >> var;
map[var] = 1 - map[var];
i = 0;
g << expresieOr();
}
return 0;
}
/*
NOT TRUE este considerat 1 ^ 1
NOT FALSE este considerat 1 ^ 0
Deaceea folosesc Xor in loc de Not
Idee de rezolvare:
Prioritizam not(1), and(2) si xor(3)
Rezolvare asemanatoare evaluare expresii de pe arhiva educationala doarc ca avem alti operatori
*/