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<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
typedef struct
{
    int s,t;
}Node;
Node e[110];
int cmp(Node u,Node v)
{
    return u.s<v.s;
}
int x[110],y[110],a[110];
int main()
{
    int n,i,j,k,s;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
        e[i].s=a[i];
        e[i].t=i+1;
    }
    sort(e,e+n,cmp);
    x[0]=e[0].t;
    x[1]=e[0].t;
    y[0]=1;
    y[1]=1;
    s=2;
    a[0]=a[0]+e[0].s+e[0].s;
    a[n]=a[n]+e[n-1].s+e[n-1].s;
    for(i=1;i<n-1;i++)
    {
        for(j=0;j<n;j++)
        {
        for(k=0;k<n;k++)
        {
            if(a[i]+e[j].s+e[k].s>=a[i-1]&&a[i]+e[j].s+e[k].s<=a[i+1])
            {
            x[s]=e[j].t;
            y[s]=i+1;
            s++;
            x[s]=e[k].t;
            y[s]=i+1;
            s++;
            break;
            }
        }
        if(k!=n)
        break;
        }
    }
    x[s]=e[n-1].t;
    y[s]=n;
    s++;
    x[s]=e[n-1].t;
    y[s]=n;
    s++;
    printf("%d\n",s);
    for(i=0;i<s;i++)
    printf("%d %d\n",x[i],y[i]);
    return 0;
}