#include<bits/stdc++.h>
using namespace std;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
#define RALL(A) A.rbegin(),A.rend()
typedef long long LL;
typedef pair<int,int> P;
const LL mod=998244353;
const LL LINF=1LL<<60;
const int INF=1<<30;
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};


int main(){
    LL n,k;cin >> n >> k;
    if(2*k-1>n){
        puts("-1");
        return 0;
    }
    vector<LL> a(n),b(n),c(n);
    int cnt = k;
    for (int i = 0; i < n; i++) {
        a[i] = cnt++;
    }
    for (int i = 0; i + n / 2 < n; i++) {
        b[n / 2 + i] = cnt++;
    }
    for (int i = 0; i < n / 2; i++) {
        b[i] = cnt++;
    }
    for (int i = 0; i + n / 2 < n; i++) {
        c[n / 2 + i] = cnt + i * 2;
    }
    for (int i = 0; i < n / 2; i++) {
        c[i] = cnt + i * 2 + 1;
    }
    for (int i = 0; i < n; i++) {
        cout << a[i] << " " << b[i] << " " << c[i] << endl;
    }
    return 0;
}

### Prompt

Generate 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;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
#define RALL(A) A.rbegin(),A.rend()
typedef long long LL;
typedef pair<int,int> P;
const LL mod=998244353;
const LL LINF=1LL<<60;
const int INF=1<<30;
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};


int main(){
    LL n,k;cin >> n >> k;
    if(2*k-1>n){
        puts("-1");
        return 0;
    }
    vector<LL> a(n),b(n),c(n);
    int cnt = k;
    for (int i = 0; i < n; i++) {
        a[i] = cnt++;
    }
    for (int i = 0; i + n / 2 < n; i++) {
        b[n / 2 + i] = cnt++;
    }
    for (int i = 0; i < n / 2; i++) {
        b[i] = cnt++;
    }
    for (int i = 0; i + n / 2 < n; i++) {
        c[n / 2 + i] = cnt + i * 2;
    }
    for (int i = 0; i < n / 2; i++) {
        c[i] = cnt + i * 2 + 1;
    }
    for (int i = 0; i < n; i++) {
        cout << a[i] << " " << b[i] << " " << c[i] << endl;
    }
    return 0;
}

```