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

const int N=1e5+10;
int a[N],b[N],c[N];

void solve(){
    int n,k;
    scanf("%d%d",&n,&k);
    if(k<=(n+1)/2){
        for(int i=1;i<=n;i++) a[i]=i+k-1;
        int index=0;
        for(int i=k+2*n-1;i>=k+n;i-=2) b[++index]=i;
        for(int i=k+2*n-2;i>=k+n;i-=2) b[++index]=i;
        index=0;
        for(int i=k+k+2*n-1;i>=k+2*n;i--) c[++index]=i;
        for(int i=k+3*n-1;i>k+k+2*n-1;i--) c[++index]=i;

        for(int i=1;i<=n;i++){
            assert(a[i]+b[i]<=c[i]);
            printf("%d %d %d\n",a[i],b[i],c[i]);
        }
    }
    else printf("-1\n");
}

int main(){

    solve();
	return 0;
}

### Prompt

In CPP, your task is to solve 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;

const int N=1e5+10;
int a[N],b[N],c[N];

void solve(){
    int n,k;
    scanf("%d%d",&n,&k);
    if(k<=(n+1)/2){
        for(int i=1;i<=n;i++) a[i]=i+k-1;
        int index=0;
        for(int i=k+2*n-1;i>=k+n;i-=2) b[++index]=i;
        for(int i=k+2*n-2;i>=k+n;i-=2) b[++index]=i;
        index=0;
        for(int i=k+k+2*n-1;i>=k+2*n;i--) c[++index]=i;
        for(int i=k+3*n-1;i>k+k+2*n-1;i--) c[++index]=i;

        for(int i=1;i<=n;i++){
            assert(a[i]+b[i]<=c[i]);
            printf("%d %d %d\n",a[i],b[i],c[i]);
        }
    }
    else printf("-1\n");
}

int main(){

    solve();
	return 0;
}

```