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: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
int acs(const void *a, const void *b) { return *(int *)a - *(int *)b; }                            /* 1,2,3,4.. */
int des(const void *a, const void *b) { return *(int *)b - *(int *)a; }                            /* 8,7,6,5.. */
int cmp_char(const void *a, const void *b) { return *(char *)a - *(char *)b; }                     /* a,b,c,d.. */
int cmp_str(const void *a, const void *b) { return strcmp(*(const char **)a, *(const char **)b); } /* aaa,aab.. */
#define min(a, b) (a < b ? a : b)
#define max(a, b) (a > b ? a : b)
#define rep(i, n) for (int i = 0; i < n; i++)
#define MAX 100001
#define MOD 1000000007

typedef long long int lli;

int main(void)
{
  int n, k;
  scanf("%d %d", &n, &k);

  int m = 0;
  int edge[MAX][2];

  // グラフを作成不可能
  int max = (n - 1) * (n - 2) / 2;
  if (k > max)
  {
    printf("-1\n");
    return 0;
  }

  // 連結グラフ(スター)を作成(頂点距離が2である個数が(n-1)(n-2)/2個)
  rep(i, n - 1)
  {
    edge[i][0] = 1;
    edge[i][1] = i + 2;
    m++;
  }

  // 頂点距離が2である個数をK個に合わせる

  rep(i, max - k)
  {
    edge[m][0] = i + 2;
    edge[m][1] = i + 3;
    m++;
  }

  printf("%d\n", m);
  rep(i, m)
  {
    printf("%d %d\n", edge[i][0], edge[i][1]);
  }
  return 0;
}