#include <bits/stdc++.h>
 
#define endl '\n'
#define fi first
#define se second
#define MOD(n,k) ( ( ((n) % (k)) + (k) ) % (k))
#define forn(i,n) for (int i = 0; i < (n); i++)
#define forr(i,a,b) for (int i = a; i <= b; i++)
#define all(v) v.begin(), v.end()
#define pb(x) push_back(x)
 
using namespace std;
 
typedef long long ll;
typedef double ld;
typedef pair<ll, ll> ii;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef vector<ii> vii;

ll n, k;
vii res;

int main () {
	ios_base::sync_with_stdio(0); cin.tie(0);
	
	cin >> n >> k;

	if (3 * n / 2 > 2 * n - k) {
		cout << -1 << endl;
		return 0;
	}

	ll i = 0, j = 3 * n / 2;

	while (i <= j) {
		if (i == j) i--;
		res.emplace_back(i, j);
		i += 2, j--;
	}

	i = 1, j = 2 * n - 1;
	while (j > 3 * n / 2) {
		res.emplace_back(i, j);
		i += 2, j--;
	}

	forn (i, res.size())
		cout << k + res[i].fi << " " << k + res[i].se << " " << k + 2 * n + i << endl;

	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 <bits/stdc++.h>
 
#define endl '\n'
#define fi first
#define se second
#define MOD(n,k) ( ( ((n) % (k)) + (k) ) % (k))
#define forn(i,n) for (int i = 0; i < (n); i++)
#define forr(i,a,b) for (int i = a; i <= b; i++)
#define all(v) v.begin(), v.end()
#define pb(x) push_back(x)
 
using namespace std;
 
typedef long long ll;
typedef double ld;
typedef pair<ll, ll> ii;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef vector<ii> vii;

ll n, k;
vii res;

int main () {
	ios_base::sync_with_stdio(0); cin.tie(0);
	
	cin >> n >> k;

	if (3 * n / 2 > 2 * n - k) {
		cout << -1 << endl;
		return 0;
	}

	ll i = 0, j = 3 * n / 2;

	while (i <= j) {
		if (i == j) i--;
		res.emplace_back(i, j);
		i += 2, j--;
	}

	i = 1, j = 2 * n - 1;
	while (j > 3 * n / 2) {
		res.emplace_back(i, j);
		i += 2, j--;
	}

	forn (i, res.size())
		cout << k + res[i].fi << " " << k + res[i].se << " " << k + 2 * n + i << endl;

	return 0;
}
```