#include <stdio.h>
#include <vector>
#include <algorithm>
#include <set>
using namespace std;
#define PB push_back
struct triple{
	int a;
	int b;
	int c;
	void out(){
		printf("%d %d %d\n",a,b,c);
		return ;
	}
};
int main(){
	int n,k,a,b,c,cnt,now=0;
	vector<triple> v;
	scanf("%d%d",&n,&k);
	v.resize(n);
	for(int i=0;i<n;i++)v[i].c=k+n*3-1-i;
	for(int i=n-1;i>=0;i-=2)v[now++].b=k+n+i;
	cnt=n-now;
	for(int i=n-2;i>=0;i-=2)v[now++].b=k+n+i;
	for(int i=cnt-1;i>=0;i--)v[--now].a=k+i;
	for(int i=n-1;i>=cnt;i--)v[--now].a=k+i;
	for(triple i:v)if(i.a+i.b>i.c){
		printf("-1\n");
		return 0;
	}
	for(triple i:v)i.out();
}

### Prompt

In cpp, your task is to solve 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 <vector>
#include <algorithm>
#include <set>
using namespace std;
#define PB push_back
struct triple{
	int a;
	int b;
	int c;
	void out(){
		printf("%d %d %d\n",a,b,c);
		return ;
	}
};
int main(){
	int n,k,a,b,c,cnt,now=0;
	vector<triple> v;
	scanf("%d%d",&n,&k);
	v.resize(n);
	for(int i=0;i<n;i++)v[i].c=k+n*3-1-i;
	for(int i=n-1;i>=0;i-=2)v[now++].b=k+n+i;
	cnt=n-now;
	for(int i=n-2;i>=0;i-=2)v[now++].b=k+n+i;
	for(int i=cnt-1;i>=0;i--)v[--now].a=k+i;
	for(int i=n-1;i>=cnt;i--)v[--now].a=k+i;
	for(triple i:v)if(i.a+i.b>i.c){
		printf("-1\n");
		return 0;
	}
	for(triple i:v)i.out();
}

```