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>

using namespace std;


int main(){
    ios_base::sync_with_stdio(0);
    int n;cin>>n;
    int ma = 0;
    int inx = -1;
    for(int i = 1;i <= n;i++) {
        int x;cin>>x;
        if(abs(x) > abs(ma)) {
            ma = x;
            inx = i;
        }
    }
    if(inx == -1) cout<<0<<endl;
    else {
        cout<<2*n-1<<endl;
        for(int i = 1;i <= n;i++) {
            cout<<inx<<' '<<i<<endl;
        }
        if(ma > 0) {
            for(int i = 1;i < n;i++) {
                cout<<i<<' '<<i+1<<endl;
            }
        } else {
            for(int i = n;i > 1;i--) {
                cout<<i<<' '<<i-1<<endl;
            }
        }
    }

    return 0;
}