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

int main(){
    int64_t N, K;
    cin >> N >> K;
    int64_t n2 = N/2;
    int64_t n1 = N-n2;
    vector<vector<int64_t>> SAB;
    for(int i=0; i<N; i++){
        int j = (i<n1 ? i+N+n2 : i+n2);
        SAB.push_back({2*K+i+j, K+i, K+j});
    }
    sort(SAB.begin(), SAB.end());
    vector<vector<int64_t>> ans(N);
    for(int i=0; i<N; i++){
        int64_t c = K + 2*N + i;
        if(SAB[i][0] > c){
            cout << -1 << endl;
            return 0;
        }
        ans[i] = {SAB[i][1], SAB[i][2], c};
    }
    for(auto& v : ans) cout << v[0] << " " << v[1] << " " << v[2] << endl;
}

### 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;

int main(){
    int64_t N, K;
    cin >> N >> K;
    int64_t n2 = N/2;
    int64_t n1 = N-n2;
    vector<vector<int64_t>> SAB;
    for(int i=0; i<N; i++){
        int j = (i<n1 ? i+N+n2 : i+n2);
        SAB.push_back({2*K+i+j, K+i, K+j});
    }
    sort(SAB.begin(), SAB.end());
    vector<vector<int64_t>> ans(N);
    for(int i=0; i<N; i++){
        int64_t c = K + 2*N + i;
        if(SAB[i][0] > c){
            cout << -1 << endl;
            return 0;
        }
        ans[i] = {SAB[i][1], SAB[i][2], c};
    }
    for(auto& v : ans) cout << v[0] << " " << v[1] << " " << v[2] << endl;
}

```