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: Accepted)

#include<bits/stdc++.h>
#define int long long
#define dd second
#define mp make_pair
#define pb push_back
#define ff first
#define dd second
#define pp pair<int,int>
using namespace std;
int n;
int tab[51];
int val = 0;
int best = 0;
int ind = 0;
main()
{
    ios_base::sync_with_stdio(false);
    cin >> n;
    for(int i = 0 ; i < n; ++i)
    {
        cin >> tab[i];
        if(abs(tab[i]) >= best)
        {
            best = abs(tab[i]);
            val = tab[i];
            ind = i;
        }
    }

    vector<pp> ans;

    for(int i = 0; i < n; ++i)
    {
        tab[i] += val;
        ans.pb(mp(ind, i));
    }

    for(int i = 0; i < n; ++i)
    {
        if(abs(tab[i]) >= best)
        {
            best = abs(tab[i]);
            val = tab[i];
            ind = i;
        }
    }

    if(val >= 0)
    {
        for(int i = 1; i < n; ++i)
        {
            //if(tab[i] < tab[i-1])
            //{
                ans.pb(mp(ind,i));
                //tab[i] += val;
                //val = tab[i];
                ind = i;
            //}
        }
    }
    else
    {
        for(int i = n-1; i >= 0; --i)
        {
            //if(tab[i] > tab[i+1])
            //{
                ans.pb(mp(ind,i));
            //    tab[i] += val;
             //   val = tab[i];
                ind = i;
            //}
        }
    }
    cout << ans.size() << "\n";
    for(auto &i:ans)
    {
        cout << i.ff+1 << " " << i.dd+1 << endl;
    }
    //for(int  i = 0; i < n; ++i) cout << tab[i] << " ";
    //cout << endl;
}