#include <stdio.h>
#include <algorithm>
#include <map>

using namespace std;

int n , k;
int a[120000] , b[120000] , c[120000] , tot;
void work () {
	int i , j , l , ti;
	scanf ( "%d%d" , &n , &k );
	if ( 2 * k - 1 > n ) {
		printf ( "-1\n" );
		return ;
	}
	for ( i = 1 , j = k ; j <= k + n - 1 ; i++ , j += 2 ) {
		a[i] = j;
	}
	ti = i - 1; l = k + n;
	for ( j -= 2 ; j >= k ; j -= 2 , ti-- ) {
		b[ti] = l;
		l++;
	}
	l = k + n + n - 1;
	for ( j = k + 1 ; j <= k + n - 1 ; i++ , j += 2 ) {
		a[i] = j;
		b[i] = l;
		l--;
	}
	for ( i = 1 ; i <= n ; i++ ) c[i] = k + n + n + i - 1;
	for ( i = 1 ; i <= n ; i++ ) {
		printf ( "%d %d %d\n" , a[i] , b[i] , c[i] );
	}
}
int main () {
	work ();
	return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

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 <stdio.h>
#include <algorithm>
#include <map>

using namespace std;

int n , k;
int a[120000] , b[120000] , c[120000] , tot;
void work () {
	int i , j , l , ti;
	scanf ( "%d%d" , &n , &k );
	if ( 2 * k - 1 > n ) {
		printf ( "-1\n" );
		return ;
	}
	for ( i = 1 , j = k ; j <= k + n - 1 ; i++ , j += 2 ) {
		a[i] = j;
	}
	ti = i - 1; l = k + n;
	for ( j -= 2 ; j >= k ; j -= 2 , ti-- ) {
		b[ti] = l;
		l++;
	}
	l = k + n + n - 1;
	for ( j = k + 1 ; j <= k + n - 1 ; i++ , j += 2 ) {
		a[i] = j;
		b[i] = l;
		l--;
	}
	for ( i = 1 ; i <= n ; i++ ) c[i] = k + n + n + i - 1;
	for ( i = 1 ; i <= n ; i++ ) {
		printf ( "%d %d %d\n" , a[i] , b[i] , c[i] );
	}
}
int main () {
	work ();
	return 0;
}

```