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 <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <set>
#include <cmath>
#include <cstring>
#include <iomanip>
#include <random>
#include <chrono>
#include <queue>
#include <ctime>
#include <cassert>
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
#define fs first
#define sc second
#define INF 1000000000
#define MOD 1000000007
#define EPS 0.00000001


int main() {
    int N, K; cin >> N >> K;
    int mx = (N-1)*(N-2)/2;

    if(K>mx){
        cout << -1 << endl;
        return 0;
    }
    
    vector<vector<bool>> E(N, vector<bool>(N));
    
    cout << mx - K + N - 1 << endl;
    
    for(int i=0; i<N-1; i++){
        cout << i+1 << " " << N << endl;
        E[i][N-1] = true;
        E[N-1][i] = true;
    }
    
    int cnt = 0;
    for(int i=0; i<N; i++){
        for(int j=0; j<N; j++){
            if(i==j) continue;
            if(cnt >= mx-K) break;
            if(!E[i][j]){
                E[i][j] = true;
                E[j][i] = true;
                cnt++;
                cout << i+1 << " " << j+1 << endl;
            }
        }
    }

    
    
    return 0;
}