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: Accepted)

/*{{{*/
#include<bits/stdc++.h>
using namespace std;

#define ll long long
#define vi vector<int>
#define vl vector<ll>
#define pii pair<int,int>
#define pll pair<ll,ll>
// pll pp; -> (pp.first,pp.second)
// vector<vl> pp(n,vl(2)); sort(pp.begin(),pp.end(),[](vl a,vl b){return a[0]<b[0];});

#define REP(i,n) for(ll (i)=0;(i)<(n);(i)++)
#define RREP(i,n) for(ll (i)=(n)-1;(i)>=0;(i)--)
#define FOR(i,a,b) for(ll (i)=(a);(i)<(b);(i)++)
#define FORR(i,a,b) for(ll (i)=(b)-1;(i)>=(a);(i)--)
#define DEBUG(x) cout << #x << ": " << x << endl
#define CHMAX(a,b) a=max((a),(b))
#define CHMIN(a,b) a=min((a),(b))
/*}}}*/
int main() {
  ll n,k;
  cin >> n >> k;
  if(k>(n-1)*(n-2)/2){
    cout << -1 << endl;
    return 0;
  }
  ll m = n*(n-1)/2-k;
  cout << m << endl;
  ll count = 0;
  REP(i,n){
    FOR(j,i+1,n){
      cout << i+1 << " " << j+1 << endl;
      if(++count==m){
        return 0;
      }
    }
  }
  return 0;
}