#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <deque>
#include <set>
#include <iomanip>
#include <utility>
#define ll long long
#define ld long double
using namespace std;

ll a[100010], b[100010], bus[100010];

int main() {
	ll N, K;
	cin >> N >> K;
	if(N<2*K-1){
		cout << -1 << endl;
		return 0;
	}
	ll t=N;
	for(int i=0; i<N; ++i){
		a[i]=t;
		++t;
		if(t==K+N) t=K;
	}
	t=K+N+(K-1)*2;
	for(int i=0; i<K; ++i){
		b[i]=t;
		bus[t-K-N]=1;
		t -= 2;
	}
	t=K+N*2-1;
	for(int i=K; i<N; ++i){
		while(bus[t-K-N]==1) --t;
		b[i]=t;
		--t;
	}
	t=K+N*3-1;
	for(int i=0; i<N; ++i){
		cout << a[i] << ' ' << b[i] << ' ' << t << endl;
		--t;
	}
	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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <deque>
#include <set>
#include <iomanip>
#include <utility>
#define ll long long
#define ld long double
using namespace std;

ll a[100010], b[100010], bus[100010];

int main() {
	ll N, K;
	cin >> N >> K;
	if(N<2*K-1){
		cout << -1 << endl;
		return 0;
	}
	ll t=N;
	for(int i=0; i<N; ++i){
		a[i]=t;
		++t;
		if(t==K+N) t=K;
	}
	t=K+N+(K-1)*2;
	for(int i=0; i<K; ++i){
		b[i]=t;
		bus[t-K-N]=1;
		t -= 2;
	}
	t=K+N*2-1;
	for(int i=K; i<N; ++i){
		while(bus[t-K-N]==1) --t;
		b[i]=t;
		--t;
	}
	t=K+N*3-1;
	for(int i=0; i<N; ++i){
		cout << a[i] << ' ' << b[i] << ' ' << t << endl;
		--t;
	}
	return 0;
}
```