#include<bits/stdc++.h>
#define rep(i, x) for(ll i = 0; i < x; i++)
#define rep2(i, x) for(ll i = 1; i <= x; i++)
#define all(a) (a).begin(),(a).end()
using ll = long long;
using ld = long double;
using namespace std;
const ll INF = 1000000000000000000;
const ll mod = 1000000007;
const ld pi = 3.141592653589793238;

signed main() {
	ll n, k, mae = 0, ato = 0; cin >> n >> k;
	ll cnt = k;
	rep(i, 2 * n) { mae += cnt; cnt++; }
	rep(i, n) { ato += cnt; cnt++; }
	if (mae > ato) { cout << "-1\n"; return 0; }

	vector<ll> v1(n), v2(n), v3(n);
	cnt = k;
	for (int i = 0; i < n; i += 2) { v1[i] = cnt; cnt++; }
	for (int i = 1; i < n; i += 2) { v1[i] = cnt; cnt++; }
	for (int i = 1; i < n; i += 2) { v2[i] = cnt; cnt++; }
	for (int i = 0; i < n; i += 2) { v2[i] = cnt; cnt++; }
	rep(i, n) { v3[i] = k + 2 * n + i; }

	rep(i, n) {
		cout << v1[i] << ' ' << v2[i] << ' ' << v3[i] << endl;
	}
	return 0;
}
### 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>
#define rep(i, x) for(ll i = 0; i < x; i++)
#define rep2(i, x) for(ll i = 1; i <= x; i++)
#define all(a) (a).begin(),(a).end()
using ll = long long;
using ld = long double;
using namespace std;
const ll INF = 1000000000000000000;
const ll mod = 1000000007;
const ld pi = 3.141592653589793238;

signed main() {
	ll n, k, mae = 0, ato = 0; cin >> n >> k;
	ll cnt = k;
	rep(i, 2 * n) { mae += cnt; cnt++; }
	rep(i, n) { ato += cnt; cnt++; }
	if (mae > ato) { cout << "-1\n"; return 0; }

	vector<ll> v1(n), v2(n), v3(n);
	cnt = k;
	for (int i = 0; i < n; i += 2) { v1[i] = cnt; cnt++; }
	for (int i = 1; i < n; i += 2) { v1[i] = cnt; cnt++; }
	for (int i = 1; i < n; i += 2) { v2[i] = cnt; cnt++; }
	for (int i = 0; i < n; i += 2) { v2[i] = cnt; cnt++; }
	rep(i, n) { v3[i] = k + 2 * n + i; }

	rep(i, n) {
		cout << v1[i] << ' ' << v2[i] << ' ' << v3[i] << endl;
	}
	return 0;
}
```