Problem: Mahmoud was trying to solve the vertex cover problem on trees. The problem statement is:

Given an undirected tree consisting of n nodes, find the minimum number of vertices that cover all the edges. Formally, we need to find a set of vertices such that for each edge (u, v) that belongs to the tree, either u is in the set, or v is in the set, or both are in the set. Mahmoud has found the following algorithm:

  * Root the tree at node 1. 
  * Count the number of nodes at an even depth. Let it be evenCnt. 
  * Count the number of nodes at an odd depth. Let it be oddCnt. 
  * The answer is the minimum between evenCnt and oddCnt. 



The depth of a node in a tree is the number of edges in the shortest path between this node and the root. The depth of the root is 0.

Ehab told Mahmoud that this algorithm is wrong, but he didn't believe because he had tested his algorithm against many trees and it worked, so Ehab asked you to find 2 trees consisting of n nodes. The algorithm should find an incorrect answer for the first tree and a correct answer for the second one.

Input

The only line contains an integer n (2 ≤ n ≤ 105), the number of nodes in the desired trees.

Output

The output should consist of 2 independent sections, each containing a tree. The algorithm should find an incorrect answer for the tree in the first section and a correct answer for the tree in the second. If a tree doesn't exist for some section, output "-1" (without quotes) for that section only.

If the answer for a section exists, it should contain n - 1 lines, each containing 2 space-separated integers u and v (1 ≤ u, v ≤ n), which means that there's an undirected edge between node u and node v. If the given graph isn't a tree or it doesn't follow the format, you'll receive wrong answer verdict.

If there are multiple answers, you can print any of them.

Examples

Input

2


Output

-1
1 2


Input

8


Output

1 2
1 3
2 4
2 5
3 6
4 7
4 8
1 2
1 3
2 4
2 5
2 6
3 7
6 8

Note

In the first sample, there is only 1 tree with 2 nodes (node 1 connected to node 2). The algorithm will produce a correct answer in it so we printed  - 1 in the first section, but notice that we printed this tree in the second section.

In the second sample:

In the first tree, the algorithm will find an answer with 4 nodes, while there exists an answer with 3 nodes like this: <image> In the second tree, the algorithm will find an answer with 3 nodes which is correct: <image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
class Point {
 public:
  double x, y;
  Point(double x = 0, double y = 0) : x(x), y(y) {}
  Point operator+(Point p) { return Point(x + p.x, y + p.y); }
  Point operator-(Point p) { return Point(x - p.x, y - p.y); }
  Point operator*(double a) { return Point(x * a, y * a); }
  Point operator/(double a) { return Point(x / a, y / a); }
  double absv() { return sqrt(norm()); }
  double norm() { return x * x + y * y; }
  bool operator<(const Point &p) const { return x != p.x ? x < p.x : y < p.y; }
  bool operator==(const Point &p) const {
    return fabs(x - p.x) < (1e-10) && fabs(y - p.y) < (1e-10);
  }
};
struct Segment {
  Point p1, p2;
};
double dot(Point a, Point b) { return a.x * b.x + a.y * b.y; }
double cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
class pa3 {
 public:
  long long x, y, z;
  pa3(long long x = 0, long long y = 0, long long z = 0) : x(x), y(y), z(z) {}
  bool operator<(const pa3 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    return z < p.z;
  }
  bool operator>(const pa3 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    return z > p.z;
  }
  bool operator==(const pa3 &p) const {
    return x == p.x && y == p.y && z == p.z;
  }
  bool operator!=(const pa3 &p) const {
    return !(x == p.x && y == p.y && z == p.z);
  }
};
bool parareru(Point a, Point b, Point c, Point d) {
  return abs(cross(a - b, d - c)) < (1e-10);
}
double distance_ls_p(Point a, Point b, Point c) {
  if (dot(b - a, c - a) < (1e-10)) return (c - a).absv();
  if (dot(a - b, c - b) < (1e-10)) return (c - b).absv();
  return abs(cross(b - a, c - a)) / (b - a).absv();
}
bool is_intersected_ls(Segment a, Segment b) {
  if (a.p1 == b.p1 || a.p2 == b.p1 || a.p1 == b.p2 || a.p2 == b.p2)
    return false;
  if (parareru((a.p2), (a.p1), (a.p1), (b.p2)) &&
      parareru((a.p2), (a.p1), (a.p1), (b.p1))) {
    if (dot(a.p1 - b.p1, a.p1 - b.p2) < (1e-10)) return true;
    if (dot(a.p2 - b.p1, a.p2 - b.p2) < (1e-10)) return true;
    if (dot(a.p1 - b.p1, a.p2 - b.p1) < (1e-10)) return true;
    if (dot(a.p1 - b.p2, a.p2 - b.p2) < (1e-10)) return true;
    return false;
  } else
    return (cross(a.p2 - a.p1, b.p1 - a.p1) * cross(a.p2 - a.p1, b.p2 - a.p1) <
            (1e-10)) &&
           (cross(b.p2 - b.p1, a.p1 - b.p1) * cross(b.p2 - b.p1, a.p2 - b.p1) <
            (1e-10));
}
double segment_dis(Segment a, Segment b) {
  if (is_intersected_ls(a, b)) return 0;
  double r = distance_ls_p(a.p1, a.p2, b.p1);
  r = min(r, distance_ls_p(a.p1, a.p2, b.p2));
  r = min(r, distance_ls_p(b.p1, b.p2, a.p2));
  r = min(r, distance_ls_p(b.p1, b.p2, a.p1));
  return r;
}
Point intersection_ls(Segment a, Segment b) {
  Point ba = b.p2 - b.p1;
  double d1 = abs(cross(ba, a.p1 - b.p1));
  double d2 = abs(cross(ba, a.p2 - b.p1));
  double t = d1 / (d1 + d2);
  return a.p1 + (a.p2 - a.p1) * t;
}
string itos(long long i) {
  ostringstream s;
  s << i;
  return s.str();
}
long long gcd(long long v, long long b) {
  if (v > b) return gcd(b, v);
  if (v == b) return b;
  if (b % v == 0) return v;
  return gcd(v, b % v);
}
double distans(double x1, double y1, double x2, double y2) {
  double rr = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2);
  return sqrt(rr);
}
long long beki(long long wa, long long rr, long long warukazu) {
  if (rr == 0) return 1ll;
  if (rr == 1) return wa % warukazu;
  if (rr % 2 == 1) return (beki(wa, rr - 1, warukazu) * wa) % warukazu;
  long long zx = beki(wa, rr / 2, warukazu);
  return (zx * zx) % warukazu;
}
map<long long, long long> ma;
long long saiki(long long r) {
  if (ma.find(r) != ma.end()) return ma[r];
  long long maa = -1;
  for (long long i = 0; i <= 55; i++)
    if (r & (1ll << i)) maa = max(maa, i);
  return ma[(1ll << maa) - 1] + saiki(r - (1ll << maa)) + (1ll << maa);
}
signed main() {
  long long n;
  cin >> n;
  if (n <= 5)
    cout << "-1" << endl;
  else {
    long long nn = n - 2;
    long long a = nn / 2;
    long long b = nn - a;
    for (long long i = 2; i <= 2 + a; i++) cout << "1 " << i << endl;
    for (long long i = 3 + a; i <= 2 + a + b; i++) cout << "2 " << i << endl;
  }
  for (long long i = 1; i <= n - 1; i++) cout << i << " " << i + 1 << endl;
  return 0;
}