#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define rrep(i,a,b) for(int i=(a);i>=(b);--i)
#define PB push_back
typedef long long LL;
const LL P = 1e9+7;
const LL INF = 1e18;
const int N = 1e6+5;
const double pi = acos(-1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int n, k;

void print(int i,int top,int diff){
	cout<<i-diff<<" "<<top-i-diff<<" "<<top-diff<<endl;
}

int main(){
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin>>n>>k;
	if(n<k*2-1){
		cout<<"-1\n";
		return 0;
	}
	if(n%2==1){
		int k1 = n+1>>1;
		int top = k1 + 3 * n - 1;
		rep(i,n,n+k1-1){
			print(i,top,k1-k);
			top--;
		}
		rep(i,k1,n-1){
			print(i,top,k1-k);
			top--;
		}
	}else{
		int k1 = n>>1;
		int top = k1 + 3 * n - 1;
		rep(i,n,n+k1-1){
			print(i,top,k1-k);
			top--;
		}
		rep(i,k1+1,n-1){
			print(i,top,k1-k);
			top--;
		}
		cout<<k<<" "<<top-n-(k1-k)<<" "<<top-(k1-k)<<endl;
	}
	return 0;
}
### Prompt

Generate a Cpp solution 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;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define rrep(i,a,b) for(int i=(a);i>=(b);--i)
#define PB push_back
typedef long long LL;
const LL P = 1e9+7;
const LL INF = 1e18;
const int N = 1e6+5;
const double pi = acos(-1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int n, k;

void print(int i,int top,int diff){
	cout<<i-diff<<" "<<top-i-diff<<" "<<top-diff<<endl;
}

int main(){
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin>>n>>k;
	if(n<k*2-1){
		cout<<"-1\n";
		return 0;
	}
	if(n%2==1){
		int k1 = n+1>>1;
		int top = k1 + 3 * n - 1;
		rep(i,n,n+k1-1){
			print(i,top,k1-k);
			top--;
		}
		rep(i,k1,n-1){
			print(i,top,k1-k);
			top--;
		}
	}else{
		int k1 = n>>1;
		int top = k1 + 3 * n - 1;
		rep(i,n,n+k1-1){
			print(i,top,k1-k);
			top--;
		}
		rep(i,k1+1,n-1){
			print(i,top,k1-k);
			top--;
		}
		cout<<k<<" "<<top-n-(k1-k)<<" "<<top-(k1-k)<<endl;
	}
	return 0;
}
```