Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
int xx[555], yy[555];
int base = 3;
int nextInt() { return base = base * 123ll % 123456721; }
int main() {
  int n, m;
  cin >> n >> m;
  if (m == 3) {
    if (n == 3) {
      printf("0 0\n3 0\n0 3\n");
      return 0;
    }
    if (n == 4) {
      printf("0 0\n100 0\n0 100\n20 20\n");
      return 0;
    }
    cout << -1 << endl;
    return 0;
  }
  double angle = 2 * acos(-1.0) / m;
  const double per = angle;
  angle = 0;
  for (int i = 0; i < m; ++i) {
    int x = (int)(99999999 * cos(angle));
    int y = (int)(99999999 * sin(angle));
    cout << x << " " << y << endl;
    xx[i] = x;
    yy[i] = y;
    angle += per;
  }
  while (true) {
    angle = 0.01;
    int shift = nextInt() % 37;
    int radius = 12368 + nextInt() % 10;
    for (int i = 0; i < n - m; ++i) {
      xx[i + m] = (int)(12368 * cos(angle)) + shift;
      yy[i + m] = (int)(12368 * sin(angle));
      angle += per;
    }
    bool checked = true;
    for (int i = 0; i < n && checked; ++i) {
      for (int j = i + 1; j < n && checked; ++j) {
        for (int k = j + 1; k < n; ++k) {
          int dx = xx[i] - xx[j];
          int dy = yy[i] - yy[j];
          int cx = xx[j] - xx[k];
          int cy = yy[j] - yy[k];
          if (dx * (long long)cy == cx * (long long)dy) {
            checked = false;
            break;
          }
        }
      }
    }
    if (checked) {
      for (int i = m; i < n; ++i) {
        cout << xx[i] << " " << yy[i] << endl;
      }
      break;
    }
  }
  return 0;
}