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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

int ans=0;
int ans1[200][2];

int main()
{
    int n;cin>>n;
    int a[100];
    int maxv=-1e9,minv=1e9;
    int maxid,minid;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        if(a[i]>maxv)
        {
            maxv=a[i];maxid=i;
        }
        if(a[i]<minv)
        {
            minv=a[i];minid=i;
        }
    }
    if(maxv>0)
    {
        if(maxv > -minv)
        {
            for(int i=2;i<=n;i++)
            {
                if(a[i]>=a[i-1])
                    continue;
                if(a[i-1]>0)
                {
                    while(a[i]<a[i-1])
                    {
                        a[i]+=a[i-1];
                        ans++;
                        ans1[ans][0]=i-1;
                        ans1[ans][1]=i;
                        if(a[i]>maxv)
                        {
                            maxv=a[i];maxid=i;
                        }
                    }
                }
                else
                {
                    a[i]+=maxv;
                    ans++;
                    ans1[ans][0]=maxid;
                    ans1[ans][1]=i;
                    if(a[i]>maxv)
                    {
                        maxv=a[i];maxid=i;
                    }
                }
            }
        }
        else
        {
            for(int i=n-1;i;i--)
            {
                if(a[i]<=a[i+1])
                    continue;
                if(a[i+1]<0)
                {
                    while(a[i]>a[i+1])
                    {
                        a[i]+=a[i+1];
                        ans++;
                        ans1[ans][0]=i+1;
                        ans1[ans][1]=i;
                        if(a[i]<minv)
                        {
                            minv=a[i];minid=i;
                        }
                    }
                }
                else
                {
                    a[i]+=minv;
                    ans++;
                    ans1[ans][0]=minid;
                    ans1[ans][1]=i;
                    if(a[i]<minv)
                    {
                        minv=a[i];minid=i;
                    }
                }
            }
        }
    }
    else
    {
            for(int i=n-1;i;i--)
            {
                if(a[i]<=a[i+1])
                    continue;
                if(a[i+1]<0)
                {
                    while(a[i]>a[i+1])
                    {
                        a[i]+=a[i+1];
                        ans++;
                        ans1[ans][0]=i+1;
                        ans1[ans][1]=i;
                        if(a[i]<minv)
                        {
                            minv=a[i];minid=i;
                        }
                    }
                }
                else
                {
                    a[i]+=minv;
                    ans++;
                    ans1[ans][0]=minid;
                    ans1[ans][1]=i;
                    if(a[i]<minv)
                    {
                        minv=a[i];minid=i;
                    }
                }
            }
    }
    cout<<ans<<endl;
    for(int i=1;i<=ans;i++)
        cout<<ans1[i][0]<<" "<<ans1[i][1]<<endl;


    return 0;
}