#include <bits/stdc++.h>
const int N = 1e5;
int a[N];

int main() {
    int n, k; scanf("%d%d", &n, &k);
    if(n < k + k - 1) puts("-1");
    else {
        int t = 0;
        for(int i = 0; i < k; ++ i) {
            printf("%d %d %d\n", k+i,  n*2+i, k+i*2+n*2);
            a[t++] = k + i*2 + n*2 + 1;    
        }
        for(int i = 0; i < t && i+k < n; ++ i) {
            printf("%d %d %d\n", i+k*2, k+n+i, a[i]);
        }
        int tmp = 3*k+n*2;
        for(int i = k*2 + t; i < n+k; ++ i) {
            printf("%d %d %d\n", i, n-k+i, tmp++);
        }
    }
   // int amo; std::cin >> amo;
}

### 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>
const int N = 1e5;
int a[N];

int main() {
    int n, k; scanf("%d%d", &n, &k);
    if(n < k + k - 1) puts("-1");
    else {
        int t = 0;
        for(int i = 0; i < k; ++ i) {
            printf("%d %d %d\n", k+i,  n*2+i, k+i*2+n*2);
            a[t++] = k + i*2 + n*2 + 1;    
        }
        for(int i = 0; i < t && i+k < n; ++ i) {
            printf("%d %d %d\n", i+k*2, k+n+i, a[i]);
        }
        int tmp = 3*k+n*2;
        for(int i = k*2 + t; i < n+k; ++ i) {
            printf("%d %d %d\n", i, n-k+i, tmp++);
        }
    }
   // int amo; std::cin >> amo;
}

```