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;

typedef long long ll;
#define REP(i,n) for(int i=0,_n=(int)(n);i<_n;++i)
#define ALL(v) (v).begin(),(v).end()
#define CLR(t,v) memset(t,(v),sizeof(t))
template<class T1,class T2>ostream& operator<<(ostream& os,const pair<T1,T2>&a){return os<<"("<<a.first<<","<<a.second<< ")";}
template<class T>void pv(T a,T b){for(T i=a;i!=b;++i)cout<<(*i)<<" ";cout<<endl;}
template<class T>void chmin(T&a,const T&b){if(a>b)a=b;}
template<class T>void chmax(T&a,const T&b){if(a<b)a=b;}


int nextInt() { int x; scanf("%d", &x); return x;}


int main2() {

  int N = nextInt();
  int K = nextInt();


  if ((N-1)*(N-2)/2 < K) {
    cout << -1 << endl;
  } else {

    bool e[N][N];
    REP(i, N) REP(j, N) e[i][j] = 0;
    REP(i, N-1) e[i][N-1] = 1;
    int k = (N-1)*(N-2)/2;
    for (int i = 0; i < N-1; i++) {
      for (int j = i+1; j < N-1; j++) {
        if (k > K) {
          k--;
          e[i][j] = e[j][i] = 1;
        }
      }
    }

    int m = 0;
    REP(i, N) for (int j = i+1; j < N; j++) if (e[i][j]) m++;
    cout << m << endl;
    for (int i = 0; i < N; i++) {
      for (int j = i+1; j < N; j++) {
        if (e[i][j]) cout << (i+1) << " " << (j+1) << endl;
      }
    }

  }
  return 0;
}

int main() {

#ifdef LOCAL
  for (;!cin.eof();cin>>ws)
#endif
    main2();
  return 0;
}