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;
int n, m;
const double PI = 3.14159265358979323846264338327;
int main() {
  cin >> n >> m;
  if (m == 3) {
    if (n == 3)
      printf("0 0\n3 0\n0 3\n");
    else if (n == 4)
      printf("0 0\n3 0\n0 3\n1 1\n");
    else
      printf("-1\n");
  } else if (m == 4) {
    pair<int, int> ans[10];
    int dx[] = {0, 1, 0, -1}, dy[] = {-1, 0, 1, 0};
    const int val = 90000000;
    for (int i = 0; i < 4; i++) {
      ans[i].first = val * cos((2.0 * PI * i) / 4);
      ans[i].second = val * sin((2.0 * PI * i) / 4);
    }
    for (int i = 0; i < 4; i++) {
      ans[i + 4].first = (ans[i].first * 0.8) + dx[i] * 100;
      ans[i + 4].second = (ans[i].second * 0.8) + dy[i] * 100;
    }
    for (int i = 0; i < n; i++) {
      printf("%d %d\n", ans[i].first, ans[i].second);
    }
  } else {
    const int val = 90000000;
    for (int i = 0; i < m; i++) {
      printf("%d %d\n", (int)(val * cos((2.0 * PI * i) / m)),
             (int)(val * sin((2.0 * PI * i) / m)));
    }
    const int val2 = val * cos((2.0 * PI) / m) + 1000;
    for (int i = 0; i < n - m; i++) {
      printf("%d %d\n", (int)(val2 * cos((2.0 * PI * i) / m)) + rand() % 1000,
             (int)(val2 * sin((2.0 * PI * i) / m)) + rand() % 1000);
    }
  }
  return 0;
}