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 (Clang 3.8.0) (judge verdict: Accepted)

#include<stdio.h>
int main()
{
	int n, k;
	scanf("%d %d", &n, &k);
	if (n == 2)
	{
		if (k == 0)
		{
			printf("1\n");
			printf("1 2\n");
			return 0;
		}
		else
		{
			printf("-1\n");
			return 0;
		}
	}
	if (n == 3)
	{
		if (k == 0)
		{
			printf("3\n");
			printf("1 2\n2 3\n1 3\n");
			return 0;
		}
		if (k == 1)
		{
			printf("2\n");
			printf("1 2\n2 3\n");
			return 0;
		}
		printf("-1\n");
		return 0;
	}
	int max = (n - 1) * (n - 2) / 2;
	if (k > max)
	{
		printf("-1\n");
		return 0;
	}
	int m = 0;
	int ans1[10004], ans2[10004];
	int i, j;
	for (i = 1; i < n; i++)
	{
		ans1[m] = 1;
		ans2[m] = i + 1;
		m++;
	}
	i = 2;
	j = 3;
	while (max > k)
	{
		max--;
		ans1[m] = i;
		ans2[m] = j;
		m++;
		j++;
		if (j > n)
		{
			i++;
			j = i + 1;
		}
	}
	printf("%d\n", m);
	for (i = 0; i < m; i++)
		printf("%d %d\n", ans1[i], ans2[i]);
	return 0;
}