//Problem solution using the stack data structure
#include <iostream>
#include <string>
#include <stack>
using namespace std;

bool match(char o, char c){ //match function it verifies if the open (o) match the closing (c)
  if((o=='(' && c==')') ||
     (o=='[' && c==']') ||
     (o=='{' && c=='}')){
  return true;
  }
  else return false;
}
int main(){
  bool bal=true; //we consider the string balanced by default
  string s="(kljfsq()([)]()()(jqs)fkl;f,sndf)";
  stack<char> ss; //We create the stack
  for (size_t i = 0; i < s.length(); i++) { //Going through the string
    if (s[i]=='('|| s[i]=='[' || s[i]=='{') { // if s[i] is a opening parenthesis or a bracket or a brace
      ss.push(s[i]); // we push it at the top of the stack
    }
    else if(s[i]==')'|| s[i]==']' || s[i]=='}'){ // if s[i] is a closing parenthesis or a bracket or a brace
      if (!ss.empty() && match(ss.top(),s[i])) { //if the stack is not empty and the top of it match s[i]
        ss.pop(); //we pop from the top of the stack
      }
      else{ //if the stack is empty or the top pf it does not match s[i]
        bal=false; //The string is not balanced (bal=false)
        break;
      }
    }
  }
  if(!ss.empty()) bal=false; //after all if the stack is not empty the string is not balanced (bal=false)
  if(bal) cout << "Balanced" << endl; //if bal is true then we print balanced
  else cout << "Not Balanced" << endl; //else we print not balanced
  return 0;
}