#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
using namespace std;
const ll mod = 1000000007;
const ll mod998 = 998244353;
const ll intmax = 2147483647;
const ll llmax = 9223372036854775807;
const char sp = ' ';

int N, K;
bool b[2000000];

int main() {
	cin >> N >> K;
	if (N < 2 * K - 1) {
		cout << -1 << endl;
		return 0;
	}
	int t = K + 2 * N - 1;
	int p = K + (N + 1) / 2 - 1;
	int s = K + N * 3 - 1;
	while (s >= K + N * 2) {
		cout << t << sp << p << sp << s << endl;
		t--;
		p--;
		s -= 2;
	}
	p = K + N - 1;
	s = K + N * 3 - 2;
	while (s >= K + N * 2) {
		cout << t << sp << p << sp << s << endl;
		t--;
		p--;
		s -= 2;
	}
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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

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

Examples

Input

1 1


Output

1 2 3


Input

3 3


Output

-1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
using namespace std;
const ll mod = 1000000007;
const ll mod998 = 998244353;
const ll intmax = 2147483647;
const ll llmax = 9223372036854775807;
const char sp = ' ';

int N, K;
bool b[2000000];

int main() {
	cin >> N >> K;
	if (N < 2 * K - 1) {
		cout << -1 << endl;
		return 0;
	}
	int t = K + 2 * N - 1;
	int p = K + (N + 1) / 2 - 1;
	int s = K + N * 3 - 1;
	while (s >= K + N * 2) {
		cout << t << sp << p << sp << s << endl;
		t--;
		p--;
		s -= 2;
	}
	p = K + N - 1;
	s = K + N * 3 - 2;
	while (s >= K + N * 2) {
		cout << t << sp << p << sp << s << endl;
		t--;
		p--;
		s -= 2;
	}
}

```