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 <bits/stdc++.h>
/// TONI BO$$ was here
/// #MLC

using namespace std;
vector < pair <int,int> > much;
int main()
{
    int n,k,i,z,a,b;
    scanf("%d%d",&n,&k);
    if(k>(n-1)*(n-2)/2)
    {
        printf("-1");
        return 0;
    }
    for(i=2; i<=n; i++)
        much.push_back({1,i});
    z=(n-1)*(n-2)/2;
    a=2;
    b=3;
    while(z>k)
    {
        much.push_back({a,b});
        b++;
        if(b>n)
            a++,b=a+1;
        z--;
    }
    printf("%d\n",much.size());
    for(i=0; i<much.size(); i++)
        printf("%d %d\n",much[i].first,much[i].second);

    return 0;
}