#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
inline int read(){
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,k;
int main(){
	n=read(),k=read();
	if(k>(n+1)/2){
		return !printf("-1\n");
	}
	int mid=(n+1)/2;
	for(int i=1;i<=mid;++i){
		printf("%d %d %d\n",k+(i-1)*2,k+n+mid-i,k+2*n+i-1);
	}
	for(int i=mid+1;i<=n;++i){
		printf("%d %d %d\n",k+(i-mid-1)*2+1,k+2*n-(i-mid),k+2*n+i-1);
	}
	return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
inline int read(){
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,k;
int main(){
	n=read(),k=read();
	if(k>(n+1)/2){
		return !printf("-1\n");
	}
	int mid=(n+1)/2;
	for(int i=1;i<=mid;++i){
		printf("%d %d %d\n",k+(i-1)*2,k+n+mid-i,k+2*n+i-1);
	}
	for(int i=mid+1;i<=n;++i){
		printf("%d %d %d\n",k+(i-mid-1)*2+1,k+2*n-(i-mid),k+2*n+i-1);
	}
	return 0;
}

```