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

#define watch(x) cerr << (#x) << ": " << (x) << endl
#define int long long

const int N = 100009;
int n, k;

int sm(int i) {
    if(i%2) {
        return (i+1)/2*i;
    } else {
        return i/2*(i+1);
    }
}

int32_t main() {

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> n >> k;
    int smsmall = sm(k+2*n-1) - sm(k-1), smbig = sm(k+3*n-1) - sm(k+2*n-1);

    if(smsmall > smbig) {
        cout << -1 << endl;
        return 0;
    }

    vector<pair<int, int>> vp;

    int r = k + 2*n - n/2 - 1;
    for(int i = 0; i < n; ++i) {
        int l = k+i;
        vp.emplace_back(l+r, l);
        r++;
        if(r == k+2*n) r = k + n;
    }

    sort(vp.begin(), vp.end());

    for(int i = 0; i < n; ++i) {
        cout << vp[i].second << ' ' << vp[i].first - vp[i].second << ' ' << k + 2*n + i <<'\n';
    }
    
    return 0;
}

### Prompt

Your task is to create 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 watch(x) cerr << (#x) << ": " << (x) << endl
#define int long long

const int N = 100009;
int n, k;

int sm(int i) {
    if(i%2) {
        return (i+1)/2*i;
    } else {
        return i/2*(i+1);
    }
}

int32_t main() {

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> n >> k;
    int smsmall = sm(k+2*n-1) - sm(k-1), smbig = sm(k+3*n-1) - sm(k+2*n-1);

    if(smsmall > smbig) {
        cout << -1 << endl;
        return 0;
    }

    vector<pair<int, int>> vp;

    int r = k + 2*n - n/2 - 1;
    for(int i = 0; i < n; ++i) {
        int l = k+i;
        vp.emplace_back(l+r, l);
        r++;
        if(r == k+2*n) r = k + n;
    }

    sort(vp.begin(), vp.end());

    for(int i = 0; i < n; ++i) {
        cout << vp[i].second << ' ' << vp[i].first - vp[i].second << ' ' << k + 2*n + i <<'\n';
    }
    
    return 0;
}

```