Problem:
Score : 100 points

Problem Statement

The door of Snuke's laboratory is locked with a security code.

The security code is a 4-digit number. We say the security code is hard to enter when it contains two consecutive digits that are the same.

You are given the current security code S. If S is hard to enter, print Bad; otherwise, print Good.

Constraints

S is a 4-character string consisting of digits.

Input

Input is given from Standard Input in the following format:

S

Output

If S is hard to enter, print Bad; otherwise, print Good.

Sample Input 1

3776

Sample Output 1

Bad

The second and third digits are the same, so 3776 is hard to enter.

Sample Input 2

8080

Sample Output 2

Good

There are no two consecutive digits that are the same, so 8080 is not hard to enter.

Sample Input 3

1333

Sample Output 3

Bad

Sample Input 4

0024

Sample Output 4

Bad

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds; 
  
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> 
 
#define FS              ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll              long long int
#define ld              long double
#define pb              push_back
#define bp              __builtin_popcount
#define sz              size()
#define ff              first
#define ss              second
#define vll             vector<ll>
#define vbool           vector<bool>
#define vpll            vector<pair<ll,ll>>
#define pll             pair<ll,ll>
#define vllv            vector<vector<ll>>
#define setpri(x)       cout<<setprecision(x)<<fixed;
#define all(v)          v.begin(),v.end()
#define allr(v)         v.rbegin(),v.rend()
#define yesr {cout<<"YES"<<endl;return;}
#define nor {cout<<"NO"<<endl;return;}
// getline (std::cin,name);
ll MOD=1e9+7;
ll ceil1(ll n,ll x){return (n-1)/x+(n>0);}

ll gcd(ll a,ll b){return __gcd(a,b);}
ll lcm(ll a,ll b){return (max(a,b)/gcd(a,b))*min(a,b);}

ll pow1(ll n,ll m ,ll mod=MOD );
ll pow2(ll n,ll k);

ll modinv(ll n,ll mod=MOD){  return pow1(n,mod-2,mod);}

bool func(pair<ll,ll> &a,pair<ll,ll> &b ){
        if(a.ff != b.ff)return a.ff < b.ff;
        return a.ss > b.ss;
}
ll const N=(ll)3e2+11;
ll const LG=(ll)log2(N)+1;


void solve(){
        ll i,j,k,l,n,m,x,y,z,r;
        
        string s;
        cin>>s;
        n=4;
        for(i=1;i<n;i++){
                if(s[i]==s[i-1]){
                        cout<<"Bad\n";
                        return;
                }
        }
        cout<<"Good\n";
        
}       
int main(){
        FS;
        
        ll i,j,k,n,m,x,y,z,q;
        q=1;
        // cin>>q;
        while(q--){
                solve();
        }
        return 0;
}


ll pow1(ll n,ll m,ll mod){
        if(m==0)return 1;
        
        if(m%2==0)return pow1((n*n)%mod,m/2,mod);
    	
    	return (pow1((n*n)%mod,m/2,mod)*n)%mod;
}

ll pow2(ll n,ll k){
        ll ans=1;
            while(k>0){
                if(k%2==1)ans=ans*n;
                n=n*n;
                k/=2;
            }
            return ans;
}