Problem:
Score : 700 points

Problem Statement

Given are positive integers N and K.

Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.

For every integer i from 1 to N, a_i + b_i \leq c_i holds.

If the answer is yes, construct one such partition.

Constraints

1 \leq N \leq 10^5

1 \leq K \leq 10^9

Input

Input is given from Standard Input in the following format:

N K

Output

If it is impossible to partition the integers satisfying the condition, print -1. If it is possible, print N triples in the following format:

a_1 b_1 c_1
:
a_N b_N c_N

Sample Input 1

1 1

Sample Output 1

1 2 3

Sample Input 2

3 3

Sample Output 2

-1

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)


#include <bits/stdc++.h>

#define M_PI       3.14159265358979323846   // pi

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
typedef tuple<ll, ll, ll> t3;
typedef tuple<ll, ll, ll, ll> t4;

#define rep(a,n) for(ll a = 0;a < n;a++)

static const ll INF = 1e18;

template<typename T>
static inline void chmin(T& ref, const T  value) {
    if (ref > value) ref = value;
}

template<typename T>
static inline void chmax(T& ref, const T value) {
    if (ref < value) ref = value;
}

const ll mod = 1000000007;

ll mpow(ll x, ll n) {
	ll ans = 1; x %= mod;
	while (n != 0) {
		if (n & 1) ans = ans * x % mod;
		x = x * x % mod;
		n = n >> 1;
	}
	return ans;
}

ll inv_mod(ll a) { return mpow(a, mod - 2); }

class Factorial {
private:
	vector<ll> fac;
	vector<ll> ifac;
public:

	Factorial(ll N) {
		fac.push_back(1);
		for (int i = 0; i < N; i++) fac.push_back(fac[i] * (i + 1) % mod);
		ifac.resize(N + 1);
		ifac[N] = inv_mod(fac[N]);
		for (int i = 0; i < N; i++) ifac[N - 1 - i] = (ifac[N - i] * (N - i)) % mod;
	}

	ll fact(ll a) { return fac[a]; }
	ll ifact(ll a) { return ifac[a]; }

	ll cmb(ll a, ll b) {
		if (a == 0 && b == 0) return 1;
		if (a < b || a < 0 || b < 0) return 0;
		ll tmp = ifact(a - b) * ifact(b) % mod;
		return tmp * fac[a] % mod;
	}
	ll per(ll a, ll b) {
		if (a == 0 && b == 0) return 1;
		if (a < b || a < 0 || b < 0) return 0;
		return fac[a] * ifac[a - b] % mod;
	}
};


int main() {
	ll n, k;
	cin >> n >> k;
	ll a = k + n - 1;
	ll b = k + 2 * n - 1;
	ll c = k + 3 * n - 1;
	if (a + b > c) {
		cout << -1 << endl;
		return 0;
	}
	rep(i, n) {
		ll a = k + i;
		ll b = k + n + i;
		ll c = k + 2 * n + i;
		cout << a <<" " <<  b << " " << c << endl;
	}
	return 0;
}