#include <bits/stdc++.h>
using namespace std;
typedef bool boolean;

const int N = 1e5 + 5;

#define pii pair<int, int>

int n, k;
pii tmp[N];
int a[N], b[N], c[N];

int main() {
	scanf("%d%d", &n, &k);
	if (2 * k - 1 > n) {
		puts("-1");
		return 0;
	}
	for (int i = 0; i < n; i++) {
		a[i] = k + i;
		c[i] = a[i] + 2 * n; 
	}
	int hn = (n + 1) >> 1;
	for (int i = 0; i < hn - i - 1; i++) {
		swap(a[i], a[hn - i - 1]);
	}
	for (int l = hn, r = n - 1; l < r; l++, r--) {
		swap(a[l], a[r]);
	}
	for (int i = 0; i < n; i++)
		tmp[i] = pii(c[i] - a[i], i);
	sort(tmp, tmp + n);
	for (int i = 0; i < n; i++)
		b[tmp[i].second] = k + n + i;
	for (int i = 0; i < n; i++) {
		printf("%d %d %d\n", a[i], b[i], c[i]);
	} 
	return 0;
}

### 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>
using namespace std;
typedef bool boolean;

const int N = 1e5 + 5;

#define pii pair<int, int>

int n, k;
pii tmp[N];
int a[N], b[N], c[N];

int main() {
	scanf("%d%d", &n, &k);
	if (2 * k - 1 > n) {
		puts("-1");
		return 0;
	}
	for (int i = 0; i < n; i++) {
		a[i] = k + i;
		c[i] = a[i] + 2 * n; 
	}
	int hn = (n + 1) >> 1;
	for (int i = 0; i < hn - i - 1; i++) {
		swap(a[i], a[hn - i - 1]);
	}
	for (int l = hn, r = n - 1; l < r; l++, r--) {
		swap(a[l], a[r]);
	}
	for (int i = 0; i < n; i++)
		tmp[i] = pii(c[i] - a[i], i);
	sort(tmp, tmp + n);
	for (int i = 0; i < n; i++)
		b[tmp[i].second] = k + n + i;
	for (int i = 0; i < n; i++) {
		printf("%d %d %d\n", a[i], b[i], c[i]);
	} 
	return 0;
}

```