#include"bits/stdc++.h"
using namespace std;
#define int long long
typedef long long ll;
#define rep(i,n) for(int i=0;i<n;i++)
typedef pair<int, int> P;
constexpr long long inf = 1e17;
constexpr int mod = 998244353;

signed main() {
	int n, k; cin >> n >> k;
	vector<int>v1(n), v2(n), v3(n);
	for (int i = 0; i < n; i += 2) {
		v1[i] = k; k++;
	}
	for (int i = 1; i < n; i += 2) {
		v1[i] = k; k++;
	}
	for (int i = 1; i < n; i += 2) {
		v2[i] = k; k++;
	}
	for (int i = 0; i < n; i += 2) {
		v2[i] = k; k++;
	}
	rep(i, n) {
		v3[i] = k; k++;
	}
	int flg = 0;
	rep(i, n) {
		if (v1[i] + v2[i] > v3[i])flg++;
	}
	if (flg)cout << -1 << endl;
	else {
		rep(i, n) {
			cout << v1[i] << ' ' << v2[i] << ' ' << v3[i] << endl;
		}
	}
}
### Prompt

In Cpp, your task is to solve 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;
#define int long long
typedef long long ll;
#define rep(i,n) for(int i=0;i<n;i++)
typedef pair<int, int> P;
constexpr long long inf = 1e17;
constexpr int mod = 998244353;

signed main() {
	int n, k; cin >> n >> k;
	vector<int>v1(n), v2(n), v3(n);
	for (int i = 0; i < n; i += 2) {
		v1[i] = k; k++;
	}
	for (int i = 1; i < n; i += 2) {
		v1[i] = k; k++;
	}
	for (int i = 1; i < n; i += 2) {
		v2[i] = k; k++;
	}
	for (int i = 0; i < n; i += 2) {
		v2[i] = k; k++;
	}
	rep(i, n) {
		v3[i] = k; k++;
	}
	int flg = 0;
	rep(i, n) {
		if (v1[i] + v2[i] > v3[i])flg++;
	}
	if (flg)cout << -1 << endl;
	else {
		rep(i, n) {
			cout << v1[i] << ' ' << v2[i] << ' ' << v3[i] << endl;
		}
	}
}
```