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>
#include <cassert>

using namespace std;
 
#define REP(i, s) for (int i = 0; i < s; ++i)
#define ALL(v) (v).begin(), (v).end()
#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl
#define EACH(i, s) for (__typeof__((s).begin()) i = (s).begin(); i != (s).end(); ++i)
#define DEBUG
#define int long long
#define INF 1e18
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<class T1, class T2> ostream& operator << (ostream &s, pair<T1,T2> P)
{ return s << '<' << P.first << ", " << P.second << '>'; }
template<class T> ostream& operator << (ostream &s, vector<T> P)
{ for (int i = 0; i < P.size(); ++i) { if (i > 0) { s << " "; } s << P[i]; } return s; }
template<class T> ostream& operator << (ostream &s, vector<vector<T> > P)
{ for (int i = 0; i < P.size(); ++i) { s << endl << P[i]; } return s << endl; }
template<class T> ostream& operator << (ostream &s, set<T> P)
{ EACH(it, P) { s << "<" << *it << "> "; } return s << endl; }
template<class T1, class T2> ostream& operator << (ostream &s, map<T1,T2> P)
{ EACH(it, P) { s << "<" << it->first << "->" << it->second << "> "; } return s << endl; }
template<class T>void show(vector<T>v){for (int i = 0; i < v.size(); i++){cerr<<v[i]<<" ";}cerr<<"\n";}
typedef long long ll;
vector<int> a;
int n;
void check()
{

    for (int i = 0; i < n - 1; i++)
    {
        assert(a[i] <= a[i + 1]);
    }
}

signed main()
{
    
    cin >> n;
    a.resize(n);
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
    }
    int max_index = 0, min_index = 0;
    int maxn = 0, minn = INF;
    for (int i = 0; i < n; i++)
    {
        if(a[i] > maxn){
            maxn = a[i];
            max_index = i;
        }
        if(a[i]<minn){
            minn = a[i];
            min_index = i;
        }
    }
    vector<pair<int, int>> v;
    if(maxn>=0 && minn<=0){
    if (abs(maxn) >= abs(minn))
    {
        for (int i = 0; i < n; i++)
        {
            v.emplace_back(max_index, i);
            a[i] += maxn;
        }
        for (int i = 0; i < n-1; i++)
        {
            if(a[i]>a[i+1]){
                v.emplace_back(i, i + 1);
                a[i + 1] += a[i];
            }
            //show(a);
        }
        //check();
    }
    else{
        for (int i = 0; i < n; i++)
        {
            v.emplace_back(min_index, i);
            a[i] += minn;
        }
        for (int i = 0; i < n; i++)
        {
            assert(a[i] <= 0);
        }

        for (int i = n-1; i >=1; i--)
        {
            if(a[i]<a[i-1]){
                v.emplace_back(i , i-1);
                a[i - 1] += a[i];
            }
           // show(a);
        }
        //show(a);
        check();
    }
    }
    else if(minn>0){
        for (int i = 0; i < n - 1; i++)
        {
            if (a[i] > a[i + 1])
            {
                v.emplace_back(i, i + 1);
                a[i + 1] += a[i];
            }
    }
    //check();
    }
    else if(maxn<0) {
        for (int i = n - 1; i >= 1; i--)
        {
            if (a[i] < a[i - 1])
            {
                v.emplace_back(i, i - 1);
                a[i - 1] += a[i];
            }
           
        }
    }

       cout << v.size() << endl;
    for (int i = 0; i < v.size(); i++)
    {
        
        cout << v[i].first + 1LL << " " << v[i].second + 1LL << endl;
    }

    return 0;
}