#include<bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N=100000;

int n,m;

void into(){
  scanf("%d%d",&n,&m);
}

int ans[N+9][3],ca;

void Get_ans(){
  if (n<2*m-1) return;
  int p0=0,p1=n-1>>1;
  for (;p0<n;p0+=2,--p1) ans[++ca][0]=m+p0,ans[ca][1]=m+n+p1,ans[ca][2]=m+n*2+ca-1;
  p0=1;p1=n-1;
  for (;p0<n;p0+=2,--p1) ans[++ca][0]=m+p0,ans[ca][1]=m+n+p1,ans[ca][2]=m+n*2+ca-1;
}

void work(){
  Get_ans();
}

void outo(){
  if (n<2*m-1) puts("-1");
  else
	for (int i=1;i<=n;++i)
	  printf("%d %d %d\n",ans[i][0],ans[i][1],ans[i][2]);
}

int main(){
  into();
  work();
  outo();
  return 0;
}
### Prompt

Your task is to create a CPP solution 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<bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N=100000;

int n,m;

void into(){
  scanf("%d%d",&n,&m);
}

int ans[N+9][3],ca;

void Get_ans(){
  if (n<2*m-1) return;
  int p0=0,p1=n-1>>1;
  for (;p0<n;p0+=2,--p1) ans[++ca][0]=m+p0,ans[ca][1]=m+n+p1,ans[ca][2]=m+n*2+ca-1;
  p0=1;p1=n-1;
  for (;p0<n;p0+=2,--p1) ans[++ca][0]=m+p0,ans[ca][1]=m+n+p1,ans[ca][2]=m+n*2+ca-1;
}

void work(){
  Get_ans();
}

void outo(){
  if (n<2*m-1) puts("-1");
  else
	for (int i=1;i<=n;++i)
	  printf("%d %d %d\n",ans[i][0],ans[i][1],ans[i][2]);
}

int main(){
  into();
  work();
  outo();
  return 0;
}
```