Problem:
Score : 600 points

Problem Statement

Snuke has an integer sequence, a, of length N. The i-th element of a (1-indexed) is a_{i}.

He can perform the following operation any number of times:

Operation: Choose integers x and y between 1 and N (inclusive), and add a_x to a_y.

He would like to perform this operation between 0 and 2N times (inclusive) so that a satisfies the condition below. Show one such sequence of operations.
It can be proved that such a sequence of operations always exists under the constraints in this problem.

Condition: a_1 \leq a_2 \leq ... \leq a_{N}

Constraints

2 \leq N \leq 50

-10^{6} \leq a_i \leq 10^{6}

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_{N}

Output

Let m be the number of operations in your solution. In the first line, print m.
In the i-th of the subsequent m lines, print the numbers x and y chosen in the i-th operation, with a space in between.
The output will be considered correct if m is between 0 and 2N (inclusive) and a satisfies the condition after the m operations.

Sample Input 1

3
-2 5 -1

Sample Output 1

2
2 3
3 3

After the first operation, a = (-2,5,4).

After the second operation, a = (-2,5,8), and the condition is now satisfied.

Sample Input 2

2
-1 -3

Sample Output 2

1
2 1

After the first operation, a = (-4,-3) and the condition is now satisfied.

Sample Input 3

5
0 0 0 0 0

Sample Output 3

0

The condition is satisfied already in the beginning.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define rrep(ri,n) for(int ri = (int)(n-1); ri >= 0; ri--)
#define rep2(i,x,n) for(int i = (int)(x); i < (int)(n); i++)
#define repit(itr,x) for(auto itr = x.begin(); itr != x.end(); itr++)
#define rrepit(ritr,x) for(auto ritr = x.rbegin(); ritr != x.rend(); ritr++)
#define ALL(x) x.begin(), x.end()
using ll = long long;
using namespace std;

void fc(vector<int> &a, int l, int r, vector<pair<int, int>> &ans){
  int mn = 1001001001, mx = -1001001001, imn = -1, imx = -1;
  rep2(i, l, r){
    if(mn > a.at(i)){
      mn = a.at(i);
      imn = i;
    }
    if(mx < a.at(i)){
      mx = a.at(i);
      imx = i;
    }
  }
  if(abs(mn) > abs(mx)){
    for(int i = imn; i > l; i--){
      ans.push_back(make_pair(i, i-1));
      ans.push_back(make_pair(i, i-1));
    }
    l = imn+1;
  }else{
    for(int i = imx; i < r-1; i++){
      ans.push_back(make_pair(i, i+1));
      ans.push_back(make_pair(i, i+1));
    }
    r = imx;
  }
  if(r - l > 0) fc(a, l, r, ans);
}

int main(){
  int n;
  cin >> n;
  vector<int> a(n);
  rep(snip_i, n) cin >> a.at(snip_i);
  vector<pair<int, int>> ans;
  fc(a, 0, n, ans);
  if(ans.size() > 2*n) return 1;
  cout << ans.size() << endl;
  for(auto p : ans){
    cout << p.first + 1 << " " << p.second + 1 << endl;
  }
  return 0;
}