#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define REP(i, n) FOR(i, 0, (n))
#define FOR(i, a, b) for(int i=(a); i<(b); i++)
#define LAR(a, b) ((a)=max((a),(b)))
#define SML(a, b) ((a)=min((a),(b)))
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pair<int, int>> vpii;
typedef pair<int, int> pii;
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define ALL(a) (a).begin(), (a).end()
#ifdef LOCAL_DEBUG
#define DEBUG(...) printf(__VA_ARGS__)
#else
#define DEBUG(...)
#endif

#define N 112345
int n;
int a[N], b[N], c[N];

int main(){
	int n, k; scanf("%d%d", &n, &k);
	if(!(k * 2 <= n+1)){
		printf("-1\n");
		return 0;
	}
	if(n == 1){
		a[0] = 1, b[0] = 2, c[0] = 3;
	} else {
		a[0] = k, a[1] = k + (n+1)/2;
		FOR(i, 2, n) a[i] = a[i-2] + 1;
		b[1] = k + n, b[0] = b[1] + n/2;
		FOR(i, 2, n) b[i] = b[i-2] + 1;
		REP(i, n) c[i] = k + 2*n + i;
	}
	REP(i, n) assert(a[i] + b[i] <= c[i]);
	REP(i, n) printf("%d %d %d\n", a[i], b[i], c[i]);
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
typedef long double ld;
#define REP(i, n) FOR(i, 0, (n))
#define FOR(i, a, b) for(int i=(a); i<(b); i++)
#define LAR(a, b) ((a)=max((a),(b)))
#define SML(a, b) ((a)=min((a),(b)))
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pair<int, int>> vpii;
typedef pair<int, int> pii;
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define ALL(a) (a).begin(), (a).end()
#ifdef LOCAL_DEBUG
#define DEBUG(...) printf(__VA_ARGS__)
#else
#define DEBUG(...)
#endif

#define N 112345
int n;
int a[N], b[N], c[N];

int main(){
	int n, k; scanf("%d%d", &n, &k);
	if(!(k * 2 <= n+1)){
		printf("-1\n");
		return 0;
	}
	if(n == 1){
		a[0] = 1, b[0] = 2, c[0] = 3;
	} else {
		a[0] = k, a[1] = k + (n+1)/2;
		FOR(i, 2, n) a[i] = a[i-2] + 1;
		b[1] = k + n, b[0] = b[1] + n/2;
		FOR(i, 2, n) b[i] = b[i-2] + 1;
		REP(i, n) c[i] = k + 2*n + i;
	}
	REP(i, n) assert(a[i] + b[i] <= c[i]);
	REP(i, n) printf("%d %d %d\n", a[i], b[i], c[i]);
}

```