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)

//Sani buyuk Osman Pasa Plevneden cikmam diyor.
//FatihSultanMehmedHan
//Abdulhamid Han
//Bismillahirrahmanirrahim
//█▀█─█──█──█▀█─█─█
//█▄█─█──█──█▄█─█▄█
//█─█─█▄─█▄─█─█─█─█

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")

#include <bits/stdc++.h>

using namespace std;

typedef long long int lo;
typedef pair< int,int > PII;

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define fio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define FOR for(lo i=1;i<=n;i++)
#define mid ((start+end)/2)
#define ort ((bas+son)/2)

const lo MAX = -1000000000000000000;
const lo MIN = 1000000000000000000;
const lo inf = 1000000000;
const lo KOK = 100000;
const lo LOG = 30;
const lo li = 500005;
const lo mod = 1000000007;

lo n,m,b[li],a[li],k,flag,t;
lo cev;
string s;
vector<int> v;

int main(void){
	scanf("%lld %lld",&n,&m);
	lo iki=(n-1)*(n-2)/2;
	if(m>iki){printf("-1\n");return 0;}
	for(lo i=2;i<=n;i++){
		printf("1 %lld\n",i);
	}
	m=iki-m;
	for(lo i=2;i<=m+1;i++){
		for(lo j=i+1;j<=m+1;j++){
			printf("%lld %lld\n",i,j);
		}
	}
	return 0;
}