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>
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
using namespace std;
typedef long long int ll;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
#define dame { puts("-1"); return 0;}
#define yn {puts("Yes");}else{puts("No");}

int main() {
    int n;
    cin >> n;
    int a[n+1];
    srep(i,1,n+1)cin >> a[i];
    
    int l = 1;
    int r = n+1;
    vp v;

    while(l<r){
        int ma = -1;
        int argma = l;
        srep(i,l,r){
            if(ma < abs(a[i])){
                ma = abs(a[i]);
                argma = i;
            }
        }
        if(ma==0)break;
        if(0<a[argma]){
            srep(i,argma+1,r){
                v.push_back(P(i-1,i));
                v.push_back(P(i-1,i));
            }
            r = argma;
        }else{
            for(int i = argma-1;l<=i;i--){
                v.push_back(P(i,i+1));
                v.push_back(P(i,i+1));
            }
            l = argma + 1;
        }
    }

    cout << v.size() << endl;
    rep(i,v.size()){
        cout << v[i].first << ' ' << v[i].second << endl;
    }

    return 0;
}