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 <iostream>
#include <vector>
#include <algorithm>
#include <map>
using namespace std;
using P = pair<int, int>;

int main(){
    int N;
    cin >> N;

    vector<int> a(N);
    int ma = -10000000;
    int mi = 10000000;
    int manum = 0;
    int minum = 0;
    for(int i = 0; i < N; ++i){
        cin >> a[i];
        if(ma < a[i]){
            ma = a[i];
            manum = i;
        }
        if(mi > a[i]){
            mi = a[i];
            minum = i;
        }
    }

    vector<P> ans;

    while(true){
        bool check = true;
        while(ma <= 0 && a[0] > mi){
            check = false;
            a[0] += mi;
            ans.push_back(P(minum + 1, 1));
        }
        if(a[0] < mi){
            mi = a[0];
            minum = 0;
        }
        for(int i = 1; i < N; ++i){
            while(a[i - 1] > a[i]){
                check = false;
                if(ma > 0){
                    a[i] += ma;
                    ans.push_back(P(manum + 1, i + 1));
                }
                else {
                    a[i - 1] += mi;
                    ans.push_back(P(minum + 1, i));
                }
            }
            if(a[i] > ma){
                ma = a[i];
                manum = i;
            }
            if(a[i - 1] < mi){
                mi = a[i - 1];
                minum = i - 1;
            }
        }
        if(check){
            break;
        }
    }

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