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

long long N, K;
long long a[100005];
long long b[100005];
vector<long long> c;

int main(){
	cin >> N >> K;

	for(long long i = K + 2 * N; i < K + 3*N; i++){
		c.push_back(i);
	}

	long long cur = 0;
	for(long long i = 0; i < N; i++){
		if(cur >= N){
			cur = 1;
		}
		//cout << "a[" << cur << "] = " <<  K + i << endl;
		a[cur] = K + i;
		cur += 2;
	}

	cur = 1;
	for(long long i = N; i < 2*N; i++){
		if(cur >= N){
			cur = 0;
		}
		//cout << "b[" << cur << "] = " << K + i << endl;
		b[cur] = K + i;
		cur += 2;
	}

	for(long long i = 0; i < N; i++){
		if(a[i] + b[i] > c[i]){
			cout << "-1" << endl;
			return 0;
		}
	}

	for(long long i = 0; i < N; i++){
		cout << a[i] << " " << b[i] << " " << c[i] << endl;
	}
}

### Prompt

Please create a solution in cpp 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;

long long N, K;
long long a[100005];
long long b[100005];
vector<long long> c;

int main(){
	cin >> N >> K;

	for(long long i = K + 2 * N; i < K + 3*N; i++){
		c.push_back(i);
	}

	long long cur = 0;
	for(long long i = 0; i < N; i++){
		if(cur >= N){
			cur = 1;
		}
		//cout << "a[" << cur << "] = " <<  K + i << endl;
		a[cur] = K + i;
		cur += 2;
	}

	cur = 1;
	for(long long i = N; i < 2*N; i++){
		if(cur >= N){
			cur = 0;
		}
		//cout << "b[" << cur << "] = " << K + i << endl;
		b[cur] = K + i;
		cur += 2;
	}

	for(long long i = 0; i < N; i++){
		if(a[i] + b[i] > c[i]){
			cout << "-1" << endl;
			return 0;
		}
	}

	for(long long i = 0; i < N; i++){
		cout << a[i] << " " << b[i] << " " << c[i] << endl;
	}
}

```