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)

// includes
#include <cstdio>
#include <cstdint>
#include <iostream>
#include <iomanip>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <algorithm>
#include <utility>
#include <functional>
#include <cmath>
#include <climits>
#include <bitset>

// macros
#define ll long long int
#define pb push_back
#define mk make_pair
#define pq priority_queue
#define FOR(i, a, b) for(int i=(a); i<(b);++i)
#define rep(i, n) FOR(i, 0, n)

using namespace std;

//  types
typedef pair<int, int> P;
typedef pair<ll, int> Pl;
typedef pair<ll, ll> Pll;
 
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;

// solve

int main(int argc, char const* argv[])
{
  int n;
  cin >> n;
  int maxi = -inf;
  int mini = inf;
  vector<int> a;
  rep(i, n){
    int x;
    cin >> x;
    a.pb(x);
    maxi = max(maxi, x);
    mini = min(mini, x);
  }
  vector<int> x, y;
  int add = 0;
  if(mini < 0 && maxi > 0){
    if(-mini > maxi)add = mini;
    else add = maxi;
    int base = 0;
    rep(i, n){
      if(add == a[i]){
        base = i;
        break;
      }
    }
    rep(i, n){
      if((ll)a[i] * (ll)add < 0){
        a[i] += add;
        x.pb(base);
        y.pb(i);
      }
    }
  }else{
    if(mini > 0)add = maxi;
    else add = mini;
  }
  int sign = (add < 0) ? - 1: 1;
  if(sign < 0){
    for(int i = n - 1; i > 0; i--){
      if(a[i-1] > a[i]){
        x.pb(i);
        y.pb(i-1);
      }
    }
  }else{
    FOR(i, 1, n){
      if(a[i-1] > a[i]){
        x.pb(i-1);
        y.pb(i);
      }
    }
  }
  cout << x.size() << endl;
  rep(i, x.size()){
    cout << x[i] + 1 << " " << y[i] + 1 << endl;
  }
	return 0;
}