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

#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
//#include<bits/stdc++.h>
#include<cstdio>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please

const int dm = 1 << 21;
char dn[dm], * di = dn;
inline void putint(int X) {
	int keta = 0;
	char C[10];
	while (X) {
		*(C + keta) = '0' + X % 10;
		X /= 10;
		keta++;
	}
	for (int i = keta - 1; i >= 0; i--)* di++ = (*(C + i));
	*di++ = '\n';
}

int main() {
	//cin.tie(0);
	//ios::sync_with_stdio(false);


	int N, K;
	scanf("%d", &N);
	scanf("%d", &K);

	if ((N + 1) / 2 >= K) {
		int A = K;
		int B = K + N + N / 2;
		int C = K + N + N;
		rep(i, (N + 1) / 2) {
			putint(A + i);
			putint(B + i);
			putint(C + i * 2);
		}
		B = K + N / 2;
		C = K + N + !(N & 1);
		for (int i = (N + 1) / 2; i < N; i++) {
			putint(A + i);
			putint(B + i);
			putint(C + i * 2);
		}
		fwrite(dn, 1, di - dn, stdout);
	}
	else printf("-1");

	Would you please return 0;
}