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 <algorithm>



using namespace std;

int main()
{
	int n, k, i, j;
	
	cin >> n >> k;
	
	if (k > (((n-1)*(n-2)) / 2))
	{
		cout << -1;
		return 0;
	}
	
	cout << (((n*(n-1)) / 2) - k) << endl;
	
	int cur_k = 0, cur_v = 1, cur_deg = n-1;
	
	while (cur_deg <= k)
	{
		k = k - cur_deg + 1;
		--cur_deg;
		++cur_v;
	}
	cur_deg-=k;
	
	for (i = 1; i < cur_v; ++i)
	{
		cout << i << " " << n << endl;
	}
	
	for (i = cur_v; i < n; ++i)
	{
		for (j = n-cur_deg+1; j <=n; ++j)
		{
			cout << i << " " << j << endl;
		}
		cur_deg = n - i - 1;
	}

	
	
	return 0;
	
}