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

int ans[100005][3];

int main(void)
{
    int n,k;
    cin >> n >> k;
    for(int i=0;i<(n+1)/2;i++,k++) ans[i*2][0] = k;
    for(int i=0;i<n/2;i++,k++) ans[i*2+1][0] = k;
    for(int i=0;i<n/2;i++,k++) ans[i*2+1][1] = k;
    for(int i=0;i<(n+1)/2;i++,k++) ans[i*2][1] = k;
    for(int i=0;i<n;i++,k++) ans[i][2] = k;
    for(int i=0;i<n;i++)
    {
        if(ans[i][0]+ans[i][1]>ans[i][2])
        {
            cout << "-1\n";
            return 0;
        }
    }
    for(int i=0;i<n;i++)
    {
        printf("%d %d %d\n",ans[i][0],ans[i][1],ans[i][2]);
    }
    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>
using namespace std;

int ans[100005][3];

int main(void)
{
    int n,k;
    cin >> n >> k;
    for(int i=0;i<(n+1)/2;i++,k++) ans[i*2][0] = k;
    for(int i=0;i<n/2;i++,k++) ans[i*2+1][0] = k;
    for(int i=0;i<n/2;i++,k++) ans[i*2+1][1] = k;
    for(int i=0;i<(n+1)/2;i++,k++) ans[i*2][1] = k;
    for(int i=0;i<n;i++,k++) ans[i][2] = k;
    for(int i=0;i<n;i++)
    {
        if(ans[i][0]+ans[i][1]>ans[i][2])
        {
            cout << "-1\n";
            return 0;
        }
    }
    for(int i=0;i<n;i++)
    {
        printf("%d %d %d\n",ans[i][0],ans[i][1],ans[i][2]);
    }
    return 0;
}
```