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 = acos(0.0) * 2.0;
const double eps = 1e-12;
const int step[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
template <class T>
inline T abs1(T a) {
  return a < 0 ? -a : a;
}
template <class T>
inline T max1(T a, T b, T c = -1234567, T d = -1234567) {
  T ans = a > b ? a : b;
  if (c != -1234567) ans = max1(ans, c);
  if (d != -1234567) ans = max1(ans, d);
  return ans;
}
template <class T>
inline T min1(T a, T b, T c = -1234567, T d = -1234567) {
  T ans = a < b ? a : b;
  if (c != -1234567) ans = min(ans, c);
  if (d != -1234567) ans = min(ans, d);
  return ans;
}
template <class T>
inline T gcd1(T a, T b) {
  if (a < b) swap(a, b);
  if (a % b == 0) return b;
  return gcd1(b, a % b);
}
template <class T>
inline T lb(T num) {
  return num & (-num);
}
inline int jud(double a, double b) {
  if (abs1(a - b) < eps) return 0;
  if (a < b) return -1;
  return 1;
}
template <class t>
inline int find(t val, t *a, int na, bool f_small = 1, bool f_lb = 1) {
  if (na == 1) return 0;
  int be = 0, en = na - 1;
  if (a[0] <= a[na - 1]) {
    if (f_lb == 0)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(a[mid], val) != 1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(a[mid], val) != -1)
          en = mid;
        else
          be = mid + 1;
      }
    if (f_small && jud(a[be], val) == 1 && be != 0) be--;
    if (!f_small && jud(a[be], val) == -1 && be != na - 1) be++;
  } else {
    if (f_lb)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(a[mid], val) != -1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(a[mid], val) != 1)
          en = mid;
        else
          be = mid + 1;
      }
    if (!f_small && jud(a[be], val) == -1 && be != 0) be--;
    if (f_small && jud(a[be], val) == 1 && be != na - 1) be++;
  }
  return be;
}
inline int bitnum(unsigned long long nValue) {
  nValue = ((0xaaaaaaaaaaaaaaaaull & nValue) >> 1) +
           (0x5555555555555555ull & nValue);
  nValue = ((0xccccccccccccccccull & nValue) >> 2) +
           (0x3333333333333333ull & nValue);
  nValue = ((0xf0f0f0f0f0f0f0f0ull & nValue) >> 4) +
           (0x0f0f0f0f0f0f0f0full & nValue);
  nValue = ((0xff00ff00ff00ff00ull & nValue) >> 8) +
           (0x00ff00ff00ff00ffull & nValue);
  nValue = ((0xffff0000ffff0000ull & nValue) >> 16) +
           (0x0000ffff0000ffffull & nValue);
  nValue = ((0xffffffff00000000ull & nValue) >> 32) +
           (0x00000000ffffffffull & nValue);
  return nValue;
}
long long pow(long long n, long long m, long long mod = 0) {
  long long ans = 1;
  long long k = n;
  while (m) {
    if (m & 1) {
      ans *= k;
      if (mod) ans %= mod;
    }
    k *= k;
    if (mod) k %= mod;
    m >>= 1;
  }
  return ans;
}
int n, m;
int main() {
  scanf("%d%d", &n, &m);
  int d = 10000000;
  if ((m == 3 && n >= 5) || (m < (n - 1) / 2 + 1) || m <= 2) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0; i < m; i++) printf("%d %d\n", i, (1 + i) * i / 2);
  for (int i = 0; i < n - m; i++) printf("%d %d\n", -i + d, (1 + i) * i / 2);
  return 0;
}