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>
#define ll long long
#define pii pair<int,int>
#define pb push_back
#define ft first
#define se second
#define mk make_pair
using namespace std;
vector<pii>ans;
int n;
ll a[105];

int gt()
{
    ll ma=a[1],pos=1;
    for(int i=2;i<=n;i++)
    {
        if(a[i]>ma)
        {
            pos=i;
            ma=a[i];
        }
    }
    return pos;
}

int gt1()
{
    ll mi=a[1],pos=1;
    for(int i=2;i<=n;i++)
    {
        if(a[i]<mi)
        {
            mi=a[i];
            pos=i;
        }
    }
    return pos;
}

int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        //scanf("%d",&a[i]);
        cin>>a[i];
    }
    
    int p=gt(),p1=gt1();
    if(abs(a[p])>abs(a[p1]))
    {
        for(int i=2;i<=n;i++)
       {
           int p=gt();

            while(a[i]<a[i-1])
           {
            ans.pb(mk(p,i));
            a[i]+=a[p];
           }

        }
    }
    else
    {
        for(int i=n-1;i>=1;i--)
        {
            int p=gt1();
            while(a[i]>a[i+1])
            {
                ans.pb(mk(p,i));
                a[i]+=a[p];
            }
        }
    }
    cout<<ans.size()<<endl;
    for(int i=0;i<ans.size();i++)
    {
        printf("%d %d\n",ans[i].ft,ans[i].se);
    }
}