#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <set>
#define ll long long
using namespace std;
template <class T>
inline void rd(T &x) {
	x=0; char c=getchar(); int f=1;
	while(!isdigit(c)) { if(c=='-') f=-1; c=getchar(); }
	while(isdigit(c)) x=x*10-'0'+c,c=getchar(); x*=f;
}
int n,k;
int vis[300010];
int main() {
	rd(n),rd(k);
	if(3*n-1-k<n/2+2*n-1) {
		printf("-1");
		return 0;
	}
	int p2=2*n-1;
	int p1=n/2;
	for(int i=3*n-1;i>=2*n;--i) {
		printf("%d %d %d\n",p1+k,p2+k,i+k);
		p2-=2;
		if(p2<n) p2=2*n-2;
		p1++;
		if(p1==n) p1=0;
	}
	return 0;
}
### Prompt

Your challenge is to write 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 <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <set>
#define ll long long
using namespace std;
template <class T>
inline void rd(T &x) {
	x=0; char c=getchar(); int f=1;
	while(!isdigit(c)) { if(c=='-') f=-1; c=getchar(); }
	while(isdigit(c)) x=x*10-'0'+c,c=getchar(); x*=f;
}
int n,k;
int vis[300010];
int main() {
	rd(n),rd(k);
	if(3*n-1-k<n/2+2*n-1) {
		printf("-1");
		return 0;
	}
	int p2=2*n-1;
	int p1=n/2;
	for(int i=3*n-1;i>=2*n;--i) {
		printf("%d %d %d\n",p1+k,p2+k,i+k);
		p2-=2;
		if(p2<n) p2=2*n-2;
		p1++;
		if(p1==n) p1=0;
	}
	return 0;
}
```