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 "bits/stdc++.h"
#include "math.h"
 
using namespace std;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<vvll> vvvll;
typedef vector<bool> vb;
typedef vector<vb> vvb;
typedef vector<int> vin;
typedef vector<pair<ll,ll> > vp;
typedef vector<vp> vvp;
#define rep(i,a,b) for(ll i=(a);i<(b);++i)
#define drep(i,a,b) for(ll i=(a);i>(b);--i)
const int MOD = 1000000007;
const int MAX = 510000;

int main(){
  ll n,k,ans=0;cin>>n>>k;
  if(k>((n-1)*(n-2))/2){cout<<-1<<endl;}
  else{
    vvb g(n,vb(n,false));
    rep(i,1,n)g[0][i]=true;
    ans=n-1;
    ll i=((n-1)*(n-2))/2-k,j=1;
    while(i>0){
      rep(k,j+1,min(n,i+j)){
        g[j][k]=true;
        ans++;
      }
      j++;
      i-=n-1-j;
    }
    cout<<ans<<endl;
    rep(k,0,n){
      rep(l,0,n){
        if(g[k][l]){cout<<k+1<<" "<<l+1<<endl;}
      }
    }
  }
}