#include <bits/stdc++.h>
using namespace std;
int n, m;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  if (m == 3) {
    if (n == 3)
      cout << 0 << ' ' << 0 << '\n' << 0 << ' ' << 1 << '\n' << 1 << ' ' << 0;
    if (n == 4)
      cout << 0 << ' ' << 0 << '\n'
           << 0 << ' ' << 3 << '\n'
           << 1 << ' ' << 1 << '\n'
           << 3 << ' ' << 0;
    if (n > 4) cout << -1;
    return 0;
  }
  for (int i = 0; i < m; i++) cout << i << ' ' << i * i + 10000000 << '\n';
  for (int i = 0; i < n - m; i++)
    cout << i << ' ' << -(i * i + 10000000) << '\n';
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

Convexity of a set of points on the plane is the size of the largest subset of points that form a convex polygon. Your task is to build a set of n points with the convexity of exactly m. Your set of points should not contain three points that lie on a straight line.

Input

The single line contains two integers n and m (3 ≤ m ≤ 100, m ≤ n ≤ 2m).

Output

If there is no solution, print "-1". Otherwise, print n pairs of integers — the coordinates of points of any set with the convexity of m. The coordinates shouldn't exceed 108 in their absolute value.

Examples

Input

4 3


Output

0 0
3 0
0 3
1 1


Input

6 3


Output

-1


Input

6 6


Output

10 0
-10 0
10 1
9 1
9 -1
0 -2


Input

7 4


Output

176166 6377
709276 539564
654734 174109
910147 434207
790497 366519
606663 21061
859328 886001

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  if (m == 3) {
    if (n == 3)
      cout << 0 << ' ' << 0 << '\n' << 0 << ' ' << 1 << '\n' << 1 << ' ' << 0;
    if (n == 4)
      cout << 0 << ' ' << 0 << '\n'
           << 0 << ' ' << 3 << '\n'
           << 1 << ' ' << 1 << '\n'
           << 3 << ' ' << 0;
    if (n > 4) cout << -1;
    return 0;
  }
  for (int i = 0; i < m; i++) cout << i << ' ' << i * i + 10000000 << '\n';
  for (int i = 0; i < n - m; i++)
    cout << i << ' ' << -(i * i + 10000000) << '\n';
}

```