#include <bits/stdc++.h>
using namespace std;
int main(){
	set<int> can;
	vector<int> stka, stkb;
	int n, k; cin >> n >> k;
	if(k > (n + 1) / 2){
		cout << -1 << endl;
		return 0;
	}
	for(int i = k; i <= k + n - 1; i++){
		can.insert(i);
		can.insert(i + n);
		can.insert(i + 2 * n);
	}
	for(int i = k; i <= k + n - 1; i++) stka.push_back(i);
	for(int i = k; i <= k + n - 1; i++) stkb.push_back(i + n);
	for(int i = k + n - 1; i >= k; i--){
		int j = i + 2 * n;
		auto doit = [&](int a, int b, int c){
			if(a + b <= c){
				if(can.count(a) && can.count(b) && can.count(c)){
					can.erase(a);
					can.erase(b);
					can.erase(c);
					cout << a << " " << b << " " << c << endl;
				}
			}
		};
		doit(j - stkb.back(), stkb.back(), j);
		doit(stka.back(), j - stka.back(), j);
		doit(stka.back(), stkb.back(), j);
		while(!stka.empty() && !can.count(stka.back())) stka.pop_back();
		while(!stkb.empty() && !can.count(stkb.back())) stkb.pop_back();
	}
	return 0;
}

### Prompt

Please formulate 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;
int main(){
	set<int> can;
	vector<int> stka, stkb;
	int n, k; cin >> n >> k;
	if(k > (n + 1) / 2){
		cout << -1 << endl;
		return 0;
	}
	for(int i = k; i <= k + n - 1; i++){
		can.insert(i);
		can.insert(i + n);
		can.insert(i + 2 * n);
	}
	for(int i = k; i <= k + n - 1; i++) stka.push_back(i);
	for(int i = k; i <= k + n - 1; i++) stkb.push_back(i + n);
	for(int i = k + n - 1; i >= k; i--){
		int j = i + 2 * n;
		auto doit = [&](int a, int b, int c){
			if(a + b <= c){
				if(can.count(a) && can.count(b) && can.count(c)){
					can.erase(a);
					can.erase(b);
					can.erase(c);
					cout << a << " " << b << " " << c << endl;
				}
			}
		};
		doit(j - stkb.back(), stkb.back(), j);
		doit(stka.back(), j - stka.back(), j);
		doit(stka.back(), stkb.back(), j);
		while(!stka.empty() && !can.count(stka.back())) stka.pop_back();
		while(!stkb.empty() && !can.count(stkb.back())) stkb.pop_back();
	}
	return 0;
}

```