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;

//C++14 (GCC 5.4.1)

#define LLINF (1LL<<60)
typedef long long ll;
#define mod 1000000007

#define repd(i,a,b) for (int i=(a);i<(b);i++)
#define rep(i,n) repd(i,0,n)
#define rrepd(i,a,b) for(int i=(a);i>=(b);i--)

template<typename T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<typename T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

template < typename T > inline string toString( const T &a ) { ostringstream oss; oss << a; return oss.str(); }

#define dbg_v1(i,V) do{cout<<"-v1----\n";rep(i,V.size())cout<<" "<<i<<"\t:"<<V[i]<<endl;cout<<endl;}while(0)
#define dbg_v2(i,k,V) do{cout<<"-v2----\n";rep(i,V.size()){rep(k,V[0].size())cout<<" "<<V[i][k];cout<<endl;}}while(0)

int main() {
    ios::sync_with_stdio(false);//stdoutとcoutの同期解除
    cin.tie(nullptr);//cinとcoutの同期解除
     
    int N,K;
    cin >>N>>K;

    ll ans=-1;
    queue<pair<ll,ll>> M;

    if(K<=(N-1)*(N-2)/2){
        for(int i=2;i<=N;i++){
          //  if(K<=0)break;
            M.push({1,i});
            K-=i-2;
          //  printf("-%d\n",K);
        }
        for(int i=2;i<N;i++){
            for(int k=i+1;k<=N;k++){
                if(K==0)break;
                M.push({i,k});
                K++;
            }
                if(K==0)break;
        }
        ans=M.size();
    }
    cout << ans <<endl;
    while(!M.empty()){
        pair<ll,ll> ne=M.front();
        cout<< ne.first  <<" "<<ne.second<<endl;   
        M.pop();
    }

    return (0);
}