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: Wrong Answer)

#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<string>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>

#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define SORT(v,n) sort(v,v+n)
#define rep(i,n) FOR(i,0,n)
#define EPS (1e-7)
#define INF (1e9)

using namespace std;
typedef long long int ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef pair<int,int> PI;
typedef pair<ll,ll> PL;
typedef priority_queue<int> PQI;
typedef priority_queue<int,vector<int>,greater<int> > PQSI;
typedef priority_queue<ll> PQL;
typedef priority_queue<ll,vector<ll>,greater<ll> > PQSL;

const int MOD = 1000000007;
struct mint { int n; mint(int n_ = 0) : n(n_) {} };
mint operator-(mint a) { return -a.n + MOD * (a.n != 0); }
mint operator+(mint a, mint b) { int x = a.n + b.n; return x - (x >= MOD) * MOD; }
mint operator-(mint a, mint b) { int x = a.n - b.n; return x + (x < 0) * MOD; }
mint operator*(mint a, mint b) { return (long long)a.n * b.n % MOD; }
mint &operator+=(mint &a, mint b) { return a = a + b; }
mint &operator-=(mint &a, mint b) { return a = a - b; }
mint &operator*=(mint &a, mint b) { return a = a * b; }
istream &operator>>(istream &i, mint &a) { return i >> a.n; }
ostream &operator<<(ostream &o, mint a) { return o << a.n; }

vector<mint> F_{1, 1}, R_{1, 1}, I_{0, 1};
void check_fact(int n) {
  for (int i = I_.size(); i <= n; i++) {
    I_.push_back(I_[MOD % i] * (MOD - MOD / i));
    F_.push_back(F_[i - 1] * i);
    R_.push_back(R_[i - 1] * I_[i]);
  }
}

mint I(int n) { check_fact(abs(n)); return n >= 0 ? I_[n] : -I_[n]; }
mint F(int n) { check_fact(n); return n < 0 ? 0 : F_[n]; }
mint R(int n) { check_fact(n); return n < 0 ? 0 : R_[n]; }
mint C(int n, int r) { return F(n) * R(n - r) * R(r); }
mint P(int n, int r) { return F(n) * R(n - r); }
mint H(int n, int r) { return n == 0 ? (r == 0) : C(n + r - 1, r); }

template<typename T>
T gcd(T x, T y){
    if (y == 0) return x;
    else return gcd(y, x % y);
}

template<typename T>
T lcm(T x, T y){
    return (x / gcd(x, y)) * y;
}

template<typename T>
T ceil(T x, T y){
    return (x + (y - 1)) / y;
}

int N, K;
int main(void){
    cin >> N >> K;
    if (K < N - 2 or C(N-1,2).n < K){
        cout << -1 << endl;
        return 0;
    }

    int r;
    for(int i = N - 1;i >= 2;i--){
        if (K > C(i,2).n){
            r = i;
            break;
        }
    }

    if (C(r,2).n+(N-r-1) != K) {
        cout << -1 << endl;
        return 0;
    }

    cout << N - 1 << endl;
    for(int i=0;i<r;i+=1) cout << 1 << " " << 2 + i << endl;
    for(int i=0;i<N-1-r;i+=1) cout << 2 + r + i - 1<< " " << r + 2 + i << endl;

    return 0;
}