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>
#define MAX_N 100

int main(void) {
  int N, K;
  scanf("%d %d", &N, &K);
  int MaxK = (N-2) * (N-1) / 2;
  if(MaxK < K) {
    printf("-1\n");
    return 0;
  }

  // Make Graph
  char D[MAX_N][MAX_N];
  D[0][0] = 0;
  for(int i = 1; i < N; i++) {
    D[0][i] = 1;
    for(int j = i; j < N; j++) {
      D[i][j] = (i == j) ? 0 : 2;
    }
  }

  int M = N-1;
  int nowK = MaxK;
  for(int i = 0; (i < N-1) && (K < nowK); i++) {
    for(int j = 1; (j < N) && (K < nowK); j++) {
      printf("i=%d j=%d, M=%d nowK=%d\n", i, j, M, nowK);
      if(D[i][j] == 2) {
	M++;
	nowK--;
	D[i][j] = 1;
      }
    }
  }
  printf("%d\n", M);
  for(int i = 0; i < N-1; i++) {
    for(int j = 1; j < N; j++) {
      if(D[i][j] == 1) {
	printf("%d %d\n", i, j);
      }
    }
  }
  return 0;
}