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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
//#include <stdlib.h>
//#include <string.h>
//#include <stdbool.h>
//#include <limits.h>
//#include <math.h>

int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  if (k>((n-1)*(n-2))/2) {
    printf("-1\n");
    return 0;
  }
  int m=n-1+((n-1)*(n-2))/2-k;
  printf("%d\n", m);
  for (int i=0; i<n-1; i++) printf("1 %d\n", i+2);
  m-=n-1;
  for (int i=1; m>0 && i<n-1; i++) {
    for (int j=i+1; m>0 && j<n; j++) {
      printf("%d %d\n", i+1, j+1);
      m--;
    }
  }
  return 0;
}