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<iostream>
    #include<vector>
    #include<cstdio>
    #include<algorithm>
    #include<map>
    using namespace std;
    typedef long long ll;
    typedef pair<ll,ll> ii;
    const int INF = 0x3f3f3f3f;
    int main()
    {
        ll n,i;
        scanf("%lld",&n);
        vector<ll> ns(n);
        for(i=0;i<n;i++)
            scanf("%lld",&ns[i]);
        ll ma=-INF,mi=INF;
        ll maw,miw;
        ll zs=0,fs=0;
        for(i=0;i<n;i++)
        {
            if(ns[i] > 0)
                zs++;
            else if(ns[i] < 0)
                fs++;
            if(ma < ns[i])
            {
                ma = ns[i];
                maw = i;
            }
            if(mi > ns[i])
            {
                mi = ns[i];
                miw = i;
            }
        }
        vector<ii> xl;
        if(ma == mi)
        {
            printf("0\n");
            return 0;
        }
        else if(mi >= 0)
        {
        ex1:for(i=1;i<n;i++)
            {
                while(ns[i-1] > ns[i])
                {
                    ii ne(maw,i);
                    xl.push_back(ne);
                    ns[i] += ns[maw];
                    if(ns[i] > ma)
                    {
                        ma = ns[i];
                        maw = i;
                    }
                }
            }
        }
        else if(ma <= 0)
        {
        ex2:for(i=n-2;i>=0;i--)
            {
                while(ns[i] > ns[i+1])
                {
                    ii ne(miw,i);
                    xl.push_back(ne);
                    ns[i] += ns[miw];
                    if(ns[i] < mi)
                    {
                        mi = ns[i];
                        miw = i;
                    }
                }
            }
        }
        else
        {
            if(-mi < ma)
                goto ex1;
            else
                goto ex2;
        }
        printf("%d\n",xl.size());
        for(i=0;i<xl.size();i++)
            printf("%lld %lld\n",xl[i].first+1,xl[i].second+1);
        return 0;
    }