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 <cstdlib>
#include <cmath>
#include <climits>
#include <cfloat>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <deque>
#include <complex>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <ctime>
#include <iterator>
#include <bitset>
#include <numeric>
#include <list>
#include <iomanip>
#include <cassert>

#if __cplusplus >= 201103L
#include <array>
#include <tuple>
#include <initializer_list>
#include <unordered_set>
#include <unordered_map>
#include <forward_list>

#define cauto const auto&
#define ALL(v) begin(v),end(v)
#else
#define ALL(v) (v).begin(),(v).end()
#endif

using namespace std;


namespace{
typedef long long LL;
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

typedef vector<int> vint;
typedef vector<vector<int> > vvint;
typedef vector<long long> vll, vLL;
typedef vector<vector<long long> > vvll, vvLL;

#define VV(T) vector<vector< T > >

template <class T>
void initvv(vector<vector<T> > &v, int a, int b, const T &t = T()){
	v.assign(a, vector<T>(b, t));
}
template <class T> inline T &chmin(T &x, const T &y){ return x = min(x, y); }
template <class T> inline T &chmax(T &x, const T &y){ return x = max(x, y); }
template <class F, class T>
void convert(const F &f, T &t){
	stringstream ss;
	ss << f;
	ss >> t;
}


#define REP(i,n) for(int i=0;i<int(n);++i)
#define RALL(v) (v).rbegin(),(v).rend()


#define MOD 1000000007LL
#define EPS 1e-8


void mainmain(){
	int n, k;
	cin >> n >> k;
	vector<pii> ans;
	for(int i = 2; i <= n && k > 0; ++i){
		ans.emplace_back(1, i);
		k -= i - 2;
	}
	if(k > 0){ cout << "-1\n"; }
	else{
		for(int i = 0; i < -k; ++i){
			ans.emplace_back(2, i + 3);
		}
		cout << ans.size() << '\n';
		for(cauto p : ans){
			cout << p.first << ' ' << p.second << '\n';
		}
	}
}



}
int main() try{
//	ios::sync_with_stdio(false); cin.tie(0);
	cout << fixed << setprecision(10);
	cerr << fixed << setprecision(4);
	mainmain();
}
catch(...){
	throw;
}