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 <algorithm>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <iomanip>
#include <iostream>
#include <string>
#include <vector>

using namespace std;

int main() {
    int n = 0, k = 0;
    cin >> n >> k;
    int a = (n-1) * (n-2) / 2;
    if (k > a) {
        cout << -1 << endl;
    }
    else {
        cout << n - 1 + a - k << endl;
        for (int i=2; i<=n; i++) {
            cout << 1 << " " << i << endl;
        }
        int now = 2;
        int nownow = 1;
        for (int i=0; i<a-k; i++) {
            cout << now << " " << now+nownow << endl;
            nownow++;
            if (now + nownow > n) {
                now++;
                nownow = 1;
            }
        }
    }
    return 0;
}