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>

#define F first
#define S second
#define PB push_back

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
typedef priority_queue<int> HEAP;
typedef priority_queue<int, vector<int>, greater<int> > RHEAP;

const int N = 100010, M = 1010;

int n, m, k;
PII res[N];

int main()
{
	cin >> n >> k;
	for (int i = 2; i <= n; i ++ ) res[m ++ ] = {i, 1};
	
	int tot = (n - 2) * (n - 1) / 2;
	if (tot < k)
	{
		puts("-1");
		return 0;
	}
	
	for (int i = 2; i <= n; i ++ )
	{
		for (int j = i + 1; j <= n; j ++ )
		{
			res[m ++ ] = {i, j};
			if ( -- tot == k) break;
		}
		if (tot == k) break;
	}
		
	cout << m << endl;
	for (int i = 0; i < m; i ++ )
		cout << res[i].F << ' ' << res[i].S << endl;
	return 0;
}