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 PI = 3.14159265358979;
const double PI2 = 6.28318530717958;
const double PId2 = 1.570796326794895;
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
T gcd(T a, T b) {
  return (b ? gcd(b, a % b) : a);
}
long long pw(long long a, long long p, int mod) {
  long long res = 1;
  while (p) {
    if (p & 1) {
      --p;
      res = (res * a) % mod;
    } else {
      p >>= 1;
      a = (a * a) % mod;
    }
  }
  return res;
}
inline vector<int> ReadVI(int count) {
  vector<int> arrayname(count);
  for (int i = 0; i < count; i++) cin >> arrayname[i];
  return arrayname;
}
inline vector<long long> ReadVlong(int count) {
  vector<long long> arrayname(count);
  for (int i = 0; i < count; i++) cin >> arrayname[i];
  return arrayname;
}
const int MOD = 1000000007;
const int MAXVALUE = 10000001;
int main() {
  int n, m;
  cin >> n >> m;
  if (m == 3) {
    if (n == 3) return printf("%d %d\n%d %d\n%d %d\n", 0, 0, 0, 1, 1, 0), 0;
    if (n == 4)
      return printf("%d %d\n%d %d\n%d %d\n%d %d", 0, 0, 0, 10, 10, 0, 1, 1), 0;
    if (n >= 5) return cout << -1 << endl, 0;
  }
  for (int i = 0; i < m; i++) printf("%d %d\n", i, i * i + 10000000);
  for (int i = 0; i < n - m; i++) printf("%d %d\n", i, -(i * i) - 10000000);
  return 0;
}