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

signed main(){
	int N, K, i;
	scanf("%lld%lld", &N, &K);
	if(2 * K > N + 1){
		printf("-1\n");
		return 0;
	}
	vector<int> a(N), b(N), c(N);
	for(i = 0; i < N; i++){
		c[i] = K + 2 * N + i;
	}
	if(N % 2 == 0){
		for(i = 0; i < N / 2; i++){
			a[i] = K + 2 * i;
		}
		for(i = 0; i < N / 2; i++){
			a[N / 2 + i] = K + 1 + 2 * i;
		}
		for(i = 0; i < N / 2; i++){
			b[i] = K + N + N / 2 - 1 - i;
		}
		for(i = 0; i < N / 2; i++){
			b[N / 2 + i] = K + N + N - 1 - i;
		}
	}
	else{
		for(i = 0; i <= N / 2; i++){
			a[i] = K + 2 * i;
		}
		for(i = 0; i < N / 2; i++){
			a[N / 2 + i + 1] = K + 1 + 2 * i;
		}
		for(i = 0; i <= N / 2; i++){
			b[i] = K + N + N / 2 - i;
		}
		for(i = 0; i < N / 2; i++){
			b[N / 2 + i + 1] = K + N + N - 1 - i;
		}
	}
	for(i = 0; i < N; i++){
		printf("%lld %lld %lld\n", a[i], b[i], c[i]);
	}
	return 0;
}
### Prompt

Please provide a CPP coded solution 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 <bits/stdc++.h>
using namespace std;
#define int long long

signed main(){
	int N, K, i;
	scanf("%lld%lld", &N, &K);
	if(2 * K > N + 1){
		printf("-1\n");
		return 0;
	}
	vector<int> a(N), b(N), c(N);
	for(i = 0; i < N; i++){
		c[i] = K + 2 * N + i;
	}
	if(N % 2 == 0){
		for(i = 0; i < N / 2; i++){
			a[i] = K + 2 * i;
		}
		for(i = 0; i < N / 2; i++){
			a[N / 2 + i] = K + 1 + 2 * i;
		}
		for(i = 0; i < N / 2; i++){
			b[i] = K + N + N / 2 - 1 - i;
		}
		for(i = 0; i < N / 2; i++){
			b[N / 2 + i] = K + N + N - 1 - i;
		}
	}
	else{
		for(i = 0; i <= N / 2; i++){
			a[i] = K + 2 * i;
		}
		for(i = 0; i < N / 2; i++){
			a[N / 2 + i + 1] = K + 1 + 2 * i;
		}
		for(i = 0; i <= N / 2; i++){
			b[i] = K + N + N / 2 - i;
		}
		for(i = 0; i < N / 2; i++){
			b[N / 2 + i + 1] = K + N + N - 1 - i;
		}
	}
	for(i = 0; i < N; i++){
		printf("%lld %lld %lld\n", a[i], b[i], c[i]);
	}
	return 0;
}
```