#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

ll n, k;
ll a[100002], b[100002], c[100002];
ll abSum, cSum;
ll minusC;

int main(){
    scanf("%lld %lld", &n, &k);
    for(ll i=k, j=1; j<=n; i++, j++){
        a[j] = i;
        abSum+=i;
    }
    for(ll i=k+n, j=1; j<=n; i++, j++){
        b[j] = i;
        abSum+=i;
    }
    for(ll i=k+n+n, j=1; j<=n; i++, j++){
        c[j] = i;
        cSum+=i;
    }
    if(abSum > cSum){
        printf("-1");
        return 0;
    }
    minusC = (cSum - abSum) / n;
    for(int i=1; i<=(n+1)/2; i++){
        printf("%lld %lld %lld\n", a[i], b[i+(n/2)], c[i*2-1]);
    }
    for(int i=(n+1)/2+1, j=1; i<=n; i++, j++){
        printf("%lld %lld %lld\n", a[i], b[j], c[2*j]);
    }
}

### Prompt

Create a solution in cpp for 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;

typedef long long ll;

ll n, k;
ll a[100002], b[100002], c[100002];
ll abSum, cSum;
ll minusC;

int main(){
    scanf("%lld %lld", &n, &k);
    for(ll i=k, j=1; j<=n; i++, j++){
        a[j] = i;
        abSum+=i;
    }
    for(ll i=k+n, j=1; j<=n; i++, j++){
        b[j] = i;
        abSum+=i;
    }
    for(ll i=k+n+n, j=1; j<=n; i++, j++){
        c[j] = i;
        cSum+=i;
    }
    if(abSum > cSum){
        printf("-1");
        return 0;
    }
    minusC = (cSum - abSum) / n;
    for(int i=1; i<=(n+1)/2; i++){
        printf("%lld %lld %lld\n", a[i], b[i+(n/2)], c[i*2-1]);
    }
    for(int i=(n+1)/2+1, j=1; i<=n; i++, j++){
        printf("%lld %lld %lld\n", a[i], b[j], c[2*j]);
    }
}

```