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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

typedef pair <int, int> ii;

int n, l;

int main()
{
    scanf("%d %d", &n, &l);
    for (int i = 2; i <= n; i++)
        for (int a = 0; i + a <= n; a++)
            for (int b = 0; i + a + b <= n; b++)
                if (i + a + b == n && i - 2 + (a + 1) * a / 2 + (b + 1) * b / 2 == l)
            {
                vector <ii> res;
                for (int z = 2; z <= i; z++)
                    res.push_back(ii(z - 1, z));
                int pnt = i + 1;
                for (int z = 0; z < a; z++)
                    res.push_back(ii(1, pnt++));
                for (int z = 0; z < b; z++)
                    res.push_back(ii(i, pnt++));
                printf("%d\n", int(res.size()));
                for (int i = 0; i < res.size(); i++)
                    printf("%d %d\n", res[i].first, res[i].second);
                return 0;
            }
    printf("-1\n");
    return 0;
}