#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define x first
#define y second
#define mp make_pair
#define pb push_back
#define sz(a) (ll)(a.size())
const ll mod = 998244353;
const ll MAX = 100005;
ll n , k;
vector<pair<ll , pair<ll , ll> > > ans;
int main()
{	
	cin >> k >> n;
	set<ll> s1 , s2;
	for(ll i = n; i < n + k; i++) s1.insert(i);
	for(ll i = n + k; i < n + 2 * k; i++) s2.insert(i);
	for(ll i = n + 2 * k; i < n + 3 * k; i++)
	{
		if(*s1.begin() <= (*s2.begin() - k))
		{
			auto it = s2.upper_bound(i - *s1.begin());
			if(it == s2.begin()){cout << -1;return 0;}
			it--;
			ans.pb({*s1.begin() , {*it , i}});
			s1.erase(*s1.begin());
			s2.erase(it);
		}
		else
		{
			auto it = s1.upper_bound(i - *s2.begin());
			if(it == s1.begin()){cout << -1;return 0;}
			it--;
			ans.pb({*s2.begin() , {*it , i}});
			s2.erase(*s2.begin());
			s1.erase(it);
		}
	}
	
	
	
	for(auto x : ans)
		cout << x.x << " " << x.y.x << " " << x.y.y << endl;
}

### Prompt

Create a solution in Cpp for 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;
typedef long long ll;
#define x first
#define y second
#define mp make_pair
#define pb push_back
#define sz(a) (ll)(a.size())
const ll mod = 998244353;
const ll MAX = 100005;
ll n , k;
vector<pair<ll , pair<ll , ll> > > ans;
int main()
{	
	cin >> k >> n;
	set<ll> s1 , s2;
	for(ll i = n; i < n + k; i++) s1.insert(i);
	for(ll i = n + k; i < n + 2 * k; i++) s2.insert(i);
	for(ll i = n + 2 * k; i < n + 3 * k; i++)
	{
		if(*s1.begin() <= (*s2.begin() - k))
		{
			auto it = s2.upper_bound(i - *s1.begin());
			if(it == s2.begin()){cout << -1;return 0;}
			it--;
			ans.pb({*s1.begin() , {*it , i}});
			s1.erase(*s1.begin());
			s2.erase(it);
		}
		else
		{
			auto it = s1.upper_bound(i - *s2.begin());
			if(it == s1.begin()){cout << -1;return 0;}
			it--;
			ans.pb({*s2.begin() , {*it , i}});
			s2.erase(*s2.begin());
			s1.erase(it);
		}
	}
	
	
	
	for(auto x : ans)
		cout << x.x << " " << x.y.x << " " << x.y.y << endl;
}

```