#include<bits/stdc++.h>

using namespace std;

#define N 1200000
#define ls (x<<1)
#define rs (x<<1|1)

int n,k,a[N],b[N],c[N];

int main(){
	scanf("%d%d",&n,&k);
	for (int i=0;i<n;++i){
		a[i]=k+i;
		b[i]=k+n+i;
		c[i]=k+n*2+i;
	}
	if (n&1){
		int x=a[0],y=b[n/2],z=c[0];
		if (x+y>z){puts("-1"); return 0;}
		if (n>1){
			x=a[n/2+1],y=b[0],z=c[1];
			if (x+y>z){puts("-1"); return 0;}
		}
		for (int i=0;i<=n/2;++i){
			int x=a[i],y=b[i+n/2],z=c[i*2];
			printf("%d %d %d\n",x,y,z);
		}
		for (int i=0;i<n/2;++i){
			int x=a[i+n/2+1],y=b[i],z=c[i*2+1];
			printf("%d %d %d\n",x,y,z);
		}
	}
	else{
		int x=a[0],y=b[n/2],z=c[0];
		if (x+y>z){puts("-1"); return 0;}
		for (int i=0;i<n/2;++i){
			int x=a[i],y=b[i+n/2],z=c[i*2];
			printf("%d %d %d\n",x,y,z);
		}
		for (int i=0;i<n/2;++i){
			int x=a[i+n/2],y=b[i],z=c[i*2+1];
			printf("%d %d %d\n",x,y,z);
		}
	}
	
	return 0;
}
			
### 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;

#define N 1200000
#define ls (x<<1)
#define rs (x<<1|1)

int n,k,a[N],b[N],c[N];

int main(){
	scanf("%d%d",&n,&k);
	for (int i=0;i<n;++i){
		a[i]=k+i;
		b[i]=k+n+i;
		c[i]=k+n*2+i;
	}
	if (n&1){
		int x=a[0],y=b[n/2],z=c[0];
		if (x+y>z){puts("-1"); return 0;}
		if (n>1){
			x=a[n/2+1],y=b[0],z=c[1];
			if (x+y>z){puts("-1"); return 0;}
		}
		for (int i=0;i<=n/2;++i){
			int x=a[i],y=b[i+n/2],z=c[i*2];
			printf("%d %d %d\n",x,y,z);
		}
		for (int i=0;i<n/2;++i){
			int x=a[i+n/2+1],y=b[i],z=c[i*2+1];
			printf("%d %d %d\n",x,y,z);
		}
	}
	else{
		int x=a[0],y=b[n/2],z=c[0];
		if (x+y>z){puts("-1"); return 0;}
		for (int i=0;i<n/2;++i){
			int x=a[i],y=b[i+n/2],z=c[i*2];
			printf("%d %d %d\n",x,y,z);
		}
		for (int i=0;i<n/2;++i){
			int x=a[i+n/2],y=b[i],z=c[i*2+1];
			printf("%d %d %d\n",x,y,z);
		}
	}
	
	return 0;
}
			
```