#include <bits/stdc++.h>
#define mset(a, b) memset(a, b, sizeof(a))
#define mcpy(a, b) memcpy(a, b, sizeof(a))
#define rg register
using namespace std;
typedef long long LL;

template <typename T> inline void read(T &AKNOI) {
    T x = 0, flag = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') flag = -1;
        ch = getchar();
    }
    while (isdigit(ch)) {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    AKNOI = flag * x;
}

int n, k;
set<int> se;
set<int>::iterator it;

void solve() {
	read(n); read(k);
	if (k * 2 > n + 1) {
		printf("-1\n");
		return;
	}
	int th = k + n * 2;
	for (int i = k + n; i < th; ++i) {
		se.insert(i);
	}
	for (int fi = k; fi < k + n; fi += 2) {
		it = se.upper_bound(th - fi);
		printf("%d %d %d\n", fi, *(--it), th);
		se.erase(it);
		++th;
	}
	for (int fi = k + 1; fi < k + n; fi += 2) {
		it = se.upper_bound(th - fi);
		printf("%d %d %d\n", fi, *(--it), th);
		se.erase(it);
		++th;
	}
}

int main() {
	solve();
	return 0;
}

### Prompt

In cpp, your task is to solve 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>
#define mset(a, b) memset(a, b, sizeof(a))
#define mcpy(a, b) memcpy(a, b, sizeof(a))
#define rg register
using namespace std;
typedef long long LL;

template <typename T> inline void read(T &AKNOI) {
    T x = 0, flag = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') flag = -1;
        ch = getchar();
    }
    while (isdigit(ch)) {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    AKNOI = flag * x;
}

int n, k;
set<int> se;
set<int>::iterator it;

void solve() {
	read(n); read(k);
	if (k * 2 > n + 1) {
		printf("-1\n");
		return;
	}
	int th = k + n * 2;
	for (int i = k + n; i < th; ++i) {
		se.insert(i);
	}
	for (int fi = k; fi < k + n; fi += 2) {
		it = se.upper_bound(th - fi);
		printf("%d %d %d\n", fi, *(--it), th);
		se.erase(it);
		++th;
	}
	for (int fi = k + 1; fi < k + n; fi += 2) {
		it = se.upper_bound(th - fi);
		printf("%d %d %d\n", fi, *(--it), th);
		se.erase(it);
		++th;
	}
}

int main() {
	solve();
	return 0;
}

```