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 <algorithm>
#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
typedef long long ll;
// const int INF = (1 << 30) - 1;
// const ll LL_INF = (1LL << 62) - 1;
// const ll MOD = 1e9 + 7;

int N, K;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> K;
  if (N < K) {
    cout << -1 << endl;
    return 0;
  }
  if (N <= 3 && K != 0) {
    cout << -1 << endl;
    return 0;
  }

  int M = N;
  vector<pair<int, int>> uv;
  for (int i = 1; i < N; i++) {
    uv.push_back(make_pair(i, i + 1));
  }
  uv.push_back(make_pair(N, 1));

  if (N > 3) {
    for (int i = 1; i <= N - K; i++) {
      if (i >= N - 1) {
        M++;
        uv.push_back(make_pair(i, (i + 2) % N));
      } else {
        M++;
        uv.push_back(make_pair(i, i + 2));
      }
    }
  }

  cout << M << endl;
  for (auto &p : uv) {
    cout << p.first << " " << p.second << endl;
  }
  return 0;
}