#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
#define int long long
int MOD = 1000000007;
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N, K;
	cin >> N >> K;
	int A1 = K;
	int A2 = K + 2 * N - 1;
	int B1 = A2 + 1;
	int B2 = K + 3 * N - 1;

	if (2 * (A1 + A2) <= B1 + B2) {

		int T = (N + 1) / 2;

		vector<int> X(N);
		vector<int> Y(N);
		vector<int> Z(N);
		for (int i = 0; i < T; i++) {
			X[i] = K + i;
			Y[i] = B1 - T + i;
			Z[i] = B1 + 2 * i;
		}
		for (int i = T; i < N; i++) {
			X[i] = K + T + i - T;
			Y[i] = A1 + N + (i - T);
			Z[i] = B1 + 1 + 2 * (i - T);
		}
		for (int i = 0; i < N; i++) {
			cout << X[i] << " " << Y[i] << " " << Z[i] << endl;
		}

	}
	else {
		cout << "-1" << endl;
	}
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
#define int long long
int MOD = 1000000007;
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N, K;
	cin >> N >> K;
	int A1 = K;
	int A2 = K + 2 * N - 1;
	int B1 = A2 + 1;
	int B2 = K + 3 * N - 1;

	if (2 * (A1 + A2) <= B1 + B2) {

		int T = (N + 1) / 2;

		vector<int> X(N);
		vector<int> Y(N);
		vector<int> Z(N);
		for (int i = 0; i < T; i++) {
			X[i] = K + i;
			Y[i] = B1 - T + i;
			Z[i] = B1 + 2 * i;
		}
		for (int i = T; i < N; i++) {
			X[i] = K + T + i - T;
			Y[i] = A1 + N + (i - T);
			Z[i] = B1 + 1 + 2 * (i - T);
		}
		for (int i = 0; i < N; i++) {
			cout << X[i] << " " << Y[i] << " " << Z[i] << endl;
		}

	}
	else {
		cout << "-1" << endl;
	}
}
```