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)

#define _USE_MATH_DEFINES
#include <array>
#include <cassert>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <algorithm>
#include <complex>
#include <cmath>
#include <numeric>
#include <bitset>
#include <functional>
#include <random>
#include <ctime>

using namespace std;

#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}

typedef long long int64;
typedef pair<int, int> ii;
const int INF = 1 << 29;
const int MOD = 1e9 + 7;
mt19937 mrand(random_device{}());
int rnd(int x) { return mrand() % x; }

int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  if (n < 2 * m - 1) {
    puts("-1");
    return 0;
  }
  vector<ii> a;
  int L = (n + 1) / 2;
  for (int i = 0, j = n - L; i < L; ++i, ++j) {
    a.emplace_back(m + i, m + n + j);
  }
  for (int i = L, j = 0; i < n; ++i, ++j) {
    a.emplace_back(m + i, m + n + j);
  }
  sort(a.begin(), a.end(), [](const ii& u, const ii& v) {
                             return u.first + u.second < v.first + v.second;
                           });
  for (int i = 0; i < n; ++i) {
    printf("%d %d %d\n", a[i].first, a[i].second, m + 2 * n + i);
  }
  return 0;
}