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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> iint;
typedef pair<ll,ll> llll;
#define ALL(x) (x).begin(),(x).end()
const ll zero = 0;
const ll INF = 3000000000000000000; //10^18
const int inINF = 1000000000; //10^9
const ll MOD = 1000000007; //10^9+7
const ll MOD2 = 998244353;

int main(){
    int N;
    cin >> N;
    vector<ll> A(N);
    for (int i = 0; i < N; i++) {
        cin >> A[i];
    }

    ll m = 0;
    int J;
    for (int i = 0; i < N; i++) {
        if(m < abs(A[i])){
            m = abs(A[i]);
            J = i;
        }
    }

    if(m == 0){
        printf("0\n");
        return 0;
    }
    printf("%d\n", 2*N);
    for (int i = 0; i < N; i++) {
        printf("%d %d\n", i+1, J+1);
    }
    if(m > 0){
        for (int i = 0; i < N; i++) {
            printf("%d %d\n", i+2, i+1);
        }
    }
    else{
        for (int i = N-1; i >= 0; i--) {
            printf("%d %d\n", i+1, i+2);
        }        
    }




}