#include <stdio.h>
#include <algorithm>
long long n, k, g;
long long a[200000];
long long b[100000];
long long c[100000];
void fp(long long x, long long y) {
	long long z = (y - x + 1) / 2;
	for (int i = 0; i < z; i++) {
		if (2 * i < z)a[x + i] = y - z / 2 + i;
		else a[x + i] = x + i + z / 2;
	}
}
int main() {
	scanf("%lld%lld", &n, &k);
	if (n % 2 == 0) {
		a[0] = 2 * n - 1;
		a[2 * n - 1] = 0;
		fp(1, 2 * n - 2);
	}
	else {
		fp(0, 2 * n - 1);
	}
	for (long long i = 0; i < n; i++) {
		b[i] = (a[i] + i + 2 * k) * n + i;
		c[i] = k + 2 * n + i;
	}
	std::sort(b, b + n);
	std::sort(c, c + n);
	for (int i = 0; i < n; i++) {
		if ((b[i] / n) > c[i])g = 1;
	}
	if (g)printf("-1\n");
	else {
		for (int i = 0; i < n; i++) {
			printf("%lld %lld %lld\n", b[i] % n + k, a[b[i] % n] + k, c[i]);
		}
	}
}
### Prompt

Please create a solution in Cpp 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 <stdio.h>
#include <algorithm>
long long n, k, g;
long long a[200000];
long long b[100000];
long long c[100000];
void fp(long long x, long long y) {
	long long z = (y - x + 1) / 2;
	for (int i = 0; i < z; i++) {
		if (2 * i < z)a[x + i] = y - z / 2 + i;
		else a[x + i] = x + i + z / 2;
	}
}
int main() {
	scanf("%lld%lld", &n, &k);
	if (n % 2 == 0) {
		a[0] = 2 * n - 1;
		a[2 * n - 1] = 0;
		fp(1, 2 * n - 2);
	}
	else {
		fp(0, 2 * n - 1);
	}
	for (long long i = 0; i < n; i++) {
		b[i] = (a[i] + i + 2 * k) * n + i;
		c[i] = k + 2 * n + i;
	}
	std::sort(b, b + n);
	std::sort(c, c + n);
	for (int i = 0; i < n; i++) {
		if ((b[i] / n) > c[i])g = 1;
	}
	if (g)printf("-1\n");
	else {
		for (int i = 0; i < n; i++) {
			printf("%lld %lld %lld\n", b[i] % n + k, a[b[i] % n] + k, c[i]);
		}
	}
}
```