#include <bits/stdc++.h>
using namespace std;

#define N 100005

int n, k;
vector <int> A, B, C;

int main() {
	scanf("%d %d", &n, &k);
    if (2*k > n+1) return puts("-1"), 0;
    else {
        for (int i = k; i < k + n; i ++) A.push_back(i), B.push_back(i + n), C.push_back(i + n * 2);
        int d = (n + 1) >> 1;
        for (int i = n-d; i < n; i ++) {
            int x = A[i];
            int y = B[n-1-(i-(n-d))*2];
            int z = C[n-(i-(n-d-1))];
            printf("%d %d %d\n", x, y, z);
        }
        for (int i = 0; i < n-d; i ++) {
            int x = A[i];
            int y = B[n-2-i*2];
            int z = C[n-d-i-1];
            printf("%d %d %d\n", x, y, z);
        }
    }
	return 0;
}

### Prompt

Please provide a cpp coded solution 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 <bits/stdc++.h>
using namespace std;

#define N 100005

int n, k;
vector <int> A, B, C;

int main() {
	scanf("%d %d", &n, &k);
    if (2*k > n+1) return puts("-1"), 0;
    else {
        for (int i = k; i < k + n; i ++) A.push_back(i), B.push_back(i + n), C.push_back(i + n * 2);
        int d = (n + 1) >> 1;
        for (int i = n-d; i < n; i ++) {
            int x = A[i];
            int y = B[n-1-(i-(n-d))*2];
            int z = C[n-(i-(n-d-1))];
            printf("%d %d %d\n", x, y, z);
        }
        for (int i = 0; i < n-d; i ++) {
            int x = A[i];
            int y = B[n-2-i*2];
            int z = C[n-d-i-1];
            printf("%d %d %d\n", x, y, z);
        }
    }
	return 0;
}

```