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: Accepted)

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,sse3,sse4,popcnt,abm,mmx")

#include <bits/stdc++.h>   
#include <stdio.h>    
 
using namespace std;
     
#define F first
#define S second
#define lb lower_bound               
#define ub upper_bound
#define pb push_back
#define pf push_front    
#define ppb pop_back
#define mp make_pair
#define bpp __builtin_popcountll                                                                                        
#define sqr(x) ((x) * (x)) 
#define al 0x3F3F3F3F
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define in insert
#define ppf pop_front
#define endl '\n'
//#define int long long
 
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
 
const int mod = (int)1e9 + 7;
const int N = (int)2e5 + 123;
const ll inf = (ll)3e18 + 1;

const double pi = acos (-1.0);
const double eps = 1e-7;
const int dx[] = {0, 0, 1, 0, -1};
const int dy[] = {0, 1, 0, -1, 0};

int n, k, d[101];
vector <pii> ans;
vector <int> g[101];
bool used[101][101];
queue <int> q;

inline void boost () {                    
   ios_base :: sync_with_stdio (NULL);
   cin.tie (NULL), cout.tie (NULL);          
}                                                     

inline void Solve () {
   cin >> n >> k;
   for (int i = 2; i <= n; i ++) {
      ans.pb ({1, i});
      used[1][i] = used[i][1] = 1;
   }
   if (k > (n - 1) * (n - 2) / 2) cout << -1, exit (0);
   int val = (n - 1) * (n - 2) / 2 - k;
   for (int i = 2; i <= n; i ++) {
      if (val == 0) break;
      for (int j = i + 1; j <= n; j ++)
         if (!used[i][j]) {
            val --;
            ans.pb ({i, j});
            if (val == 0) break;
         }
   }
   cout << sz (ans) << endl;
   for (auto it : ans) cout << it.F << ' ' << it.S << endl;


}                                     

main () {                                       
// freopen (".in", "r", stdin);    
// freopen (".out", "w", stdout);              
   boost ();
   int tt = 1;
// cin >> tt;   
   while (tt --) {
      Solve ();
   }                                               
   return 0;      
}