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>
using namespace std;

//K = 0 のとき、完全グラフ
//辺を n 本取り除くと、ペア増える (要検証)
//max pair is (n - 1)(n - 2) / 2
//Σ(i = 0 to n)(頂点iの次数 C 2) = ペア数

int main(){
    int N, K;
    cin >> N >> K;
    int M = (N - 1) * (N - 2) / 2;
    if(M <= K){
        cout << -1 << endl;
    }else{
        vector<pair<int, int>> ans;
        for(int i = 1; i <= N - 1; i++){
            for(int j = i + 1; j <= N; j++){
                ans.push_back(make_pair(i, j));
            }
        }
        int max_E = N * (N - 1) / 2;
        int Edge = max_E - K;
        cout << Edge << endl;
        for(int i = 0; i < Edge; i++){
            cout << ans[i].first << " " << ans[i].second << endl;
        }
    }
    return 0;
}