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

struct node {
	int s, a, b;
	node(int a = 0, int b = 0) : a(a), b(b) {
		s = a + b;
	}
	bool operator < (const node& rhs) const { return s < rhs.s; }
} a[100050];

int main() {
	int n, k;
	scanf("%d%d", &n, &k);
	int val = k * 2 + 2 * n - 1;
	int m = n / 2;
	for(int i = 0; i < m; i++) a[i] = node(k + i, k + n + i + n - m);
	for(int i = m; i < n; i++) a[i] = node(k + i, k + n + i - m);
	sort(a, a + n);
	bool ans = true;
	for(int i = 0; i < n; i++)
		if(a[i].s > k + 2 * n + i) ans = false;
	if(!ans) {
		puts("-1"); return 0;
	}
	for(int i = 0; i < n; i++) printf("%d %d %d\n", a[i].a, a[i].b, k + 2 * n + i);
	return 0;
}

### Prompt

Please formulate 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;

struct node {
	int s, a, b;
	node(int a = 0, int b = 0) : a(a), b(b) {
		s = a + b;
	}
	bool operator < (const node& rhs) const { return s < rhs.s; }
} a[100050];

int main() {
	int n, k;
	scanf("%d%d", &n, &k);
	int val = k * 2 + 2 * n - 1;
	int m = n / 2;
	for(int i = 0; i < m; i++) a[i] = node(k + i, k + n + i + n - m);
	for(int i = m; i < n; i++) a[i] = node(k + i, k + n + i - m);
	sort(a, a + n);
	bool ans = true;
	for(int i = 0; i < n; i++)
		if(a[i].s > k + 2 * n + i) ans = false;
	if(!ans) {
		puts("-1"); return 0;
	}
	for(int i = 0; i < n; i++) printf("%d %d %d\n", a[i].a, a[i].b, k + 2 * n + i);
	return 0;
}

```