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>
using namespace std;
template<class T> inline bool chmax(T &a, T b){
    if(a < b){
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmin(T &a, T b){
    if(a > b){
        a = b;
        return true;
    }
    return false;
}

int main(){
    int N;
    long long Mi = 1e9, Ma = -1e9;
    int mi, ma;
    cin >> N;
    vector<long long> A(N);
    for(int i = 0; i < N; i++){
        cin >> A[i];
        if(chmin(Mi, A[i]))mi = i + 1;
        if(chmax(Ma, A[i]))ma = i + 1;
    }
    
    int ans = 0;
    vector<pair<int, int>> anss;
    if(Mi >= 0){
        for(int i = 1; i < N; i++){
            while(A[i - 1] > A[i]){
                A[i] += Ma;
                ans++;
                anss.push_back(make_pair(ma, i + 1));
            }
        }
    }else{
        if(Ma <= 0){
            for(int i = 1; i < N; i++){
                while(A[i - 1] > A[i]){
                    A[i - 1] += Mi;
                    ans++;
                    anss.push_back(make_pair(mi, i));
                }
            }
        }else{
            if(abs(Ma) < abs(Mi)){
                for(int i = 1; i < N; i++){
                    while(A[i - 1] > A[i]){
                        A[i - 1] += Mi;
                        ans++;
                        anss.push_back(make_pair(mi, i));
                    }
                }
            }else{
                for(int i = 1; i < N; i++){
                    while(A[i - 1] > A[i]){
                        A[i] += Ma;
                        ans++;
                        anss.push_back(make_pair(ma, i + 1));
                    }
                }
            }
        }
    }
    
    cout << ans << endl;
    for(int i = 0; i < anss.size(); i++){
        cout << anss[i].first << " " << anss[i].second << endl;
    }
    return 0;
}