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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
#include<sstream>
#include<iomanip>
#include<limits>
#include<deque>
#include<map>
#include<list>
#include<set>
#include <unordered_set>
#include<vector>
#include<cmath>
#include<cstdio>
#include<memory>
#include<bitset>
#include<stack>
#include<functional>
#include<queue>
#include<regex>
#include<time.h>

using namespace std;


#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl
#define yes cout<<"yes"<<endl
#define no cout<<"no"<<endl

#define Tof(x) (x) ? Yes : No
#define TOF(x) (x) ? YES : NO
#define tof(x) (x) ? yes : no

using ll = long long;

constexpr ll MOD = 1000000007;
constexpr ll INF = 1050000000;


/*vectorの要素を全部確認するやつだよ。普通に拡張for文したほうがいいよ。

for (vector<int>::iterator itr = ans.begin(); itr != ans.end(); ++itr) {
	cout << *itr << " ";
}

*/


/*for文でx,yを上下左右確認するやつだよ。正直使いづらいよ。

int dy[] = { 0, 1, 0, -1 };
int dx[] = { 1, 0, -1, 0 };


for (int i = 0; i < 4; i++) {
	int ny = y + dy[i];
	int nx = x + dx[i];
}

*/


/*繰り返し二乗法だよ。modもとってくれるよ。n^kをmodで割った余りでやってくれるよ。

ll POW_MOD(ll n, ll k, ll mod) {
	ll r = 1;

	for (; k > 0; k >>= 1) {
		if (k & 1) {
			r = (r * n) % mod;
		}
		n = (n * n) % mod;
	}
	return r;
}

*/


/*実行開始からの経過時間を表すよ。

while (clock() < 1900000) {  }

*/


/*まだ使えないテクニックのメモだよ。

for (int bit = 0; bit < (1<<n); ++bit)//bit全探索用のfor文だよ

*/


/*2重for文書くのめんどくなった時用だよ。

for (int i = 0; i < h; ++i) {
	for (int j = 0; j < w; ++j) {

	}
}

*/

int main(){

	string s; cin >> s;

	if (s[0] == s[1] || s[2] == s[1] || s[2] == s[3]) { cout << "Bad" << endl; }
	else {
		cout << "Good" << endl;
	}


	return 0;
}