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<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<map>
#include<set>
#include<cstdio>
#include<cmath>
#include<numeric>
#include<queue>
#include<stack>
#define repp(i,a,b) for(int i=(int)a;i<(int)b;++i)
#define repm(i,a,b) for(int i=(int)a;i>(int)b;--i)
using ll=long long;
static const ll mod = 1e9 + 7;
static const ll INF = 1LL << 50;
using namespace std;


int main() {
	//スターグラフにK本足すか、完全グラフからK本引く
	int n, k;
	cin >> n >> k;
	bool loop = true;
	if (k > (n - 1) * (n - 2) / 2) {
		cout << -1 << endl;
		return 0;
	}
	vector<pair<int, int>>v;
	repp(i, 1, n) {
		//ウニグラフを作る
		v.emplace_back(0, i);
	}int cnt = 0;
	repp(j, 1, n) {
		if (!loop)break;
		repp(l, j + 1, n) {
			if (l== j)continue;
			v.emplace_back(j, l);
			++cnt;
			if (cnt == (n - 1) * (n - 2) / 2 - k) {
				loop = false;
				break;
			}
		}
	}
			cout << n * (n - 1) / 2 - k << endl;
			repp(i, 0, v.size())cout << v[i].first + 1 << " " << v[i].second + 1 << endl;
			return 0;
}