Problem:
Score: 500 points

Problem Statement

Does there exist an undirected graph with N vertices satisfying the following conditions?

The graph is simple and connected.

The vertices are numbered 1, 2, ..., N.

Let M be the number of edges in the graph. The edges are numbered 1, 2, ..., M, the length of each edge is 1, and Edge i connects Vertex u_i and Vertex v_i.

There are exactly K pairs of vertices (i,\ j)\ (i < j) such that the shortest distance between them is 2.

If there exists such a graph, construct an example.

Constraints

All values in input are integers.

2 \leq N \leq 100

0 \leq K \leq \frac{N(N - 1)}{2}

Input

Input is given from Standard Input in the following format:

N K

Output

If there does not exist an undirected graph with N vertices satisfying the conditions, print -1.

If there exists such a graph, print an example in the following format (refer to Problem Statement for what the symbols stand for):

M
u_1 v_1
:
u_M v_M

If there exist multiple graphs satisfying the conditions, any of them will be accepted.

Sample Input 1

5 3

Sample Output 1

5
4 3
1 2
3 1
4 5
2 3

This graph has three pairs of vertices such that the shortest distance between them is 2: (1,\ 4), (2,\ 4), and (3,\ 5). Thus, the condition is satisfied.

Sample Input 2

5 8

Sample Output 2

-1

There is no graph satisfying the conditions.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
#include <set>
#include <queue>
#include <map>
#include <string>

#define rep(i, a, b) for ( int i = (a); i < (b); i++ )
#define per(i, a, b) for ( int i = (b)-1; i >= (a); i--)
#define pb push_back
#define mp make_pair
#define bg begin()
#define en end()
#define all(v) (v).begin(), (v).end()
#define sz(v) (int)(v).size()

using namespace std;

typedef long long ll;
typedef long double ld;

static const long long MOD = 998244353;


int n, k;
vector<pair<int, int> > vec;

int main(void) {
  cin >> n >> k;
  
  for (int i=2; i<=n; i++) vec.pb(mp(1, i));
  int t = ((n-1)*(n-2))/2;
  
  int u = 2, v = u + 1;
  while (t>k) {
    vec.pb(mp(u, v));
    t--;
    v++;
    if (v>u) {
      u++;
      v = u + 1;
    }
  
  }
  
  if (t!=k) puts("-1");
  else for (auto p : vec) printf("%d %d\n", p.first, p.second);

    return 0;
}