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)

#define _USE_MATH_DEFINES
#include <cstdio>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <complex>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <limits>
#include <climits>
#include <cfloat>
#include <functional>
#include <iterator>
using namespace std;

void solve1(vector<long long> v, vector<pair<int, int> >& ans)
{
    int n = v.size();
    ans.clear();

    for(;;){
        int x = max_element(v.begin(), v.end()) - v.begin();
        int y = 1;
        while(y < n && v[y-1] <= v[y])
            ++ y;
        if(y == n)
            return;
        v[y] += v[x];
        ans.push_back(make_pair(x, y));
    }
}

void solve2(vector<long long> v, vector<pair<int, int> >& ans)
{
    int n = v.size();
    ans.clear();

    for(;;){
        int x = min_element(v.begin(), v.end()) - v.begin();
        int y = n - 2;
        while(y >= 0 && v[y] <= v[y+1])
            -- y;
        if(y == -1)
            return;
        v[y] += v[x];
        ans.push_back(make_pair(x, y));
    }
}

void solve(const vector<long long>& v, vector<pair<int, int> >& ans)
{
    if(-*min_element(v.begin(), v.end()) < *max_element(v.begin(), v.end()))
        solve1(v, ans);
    else
        solve2(v, ans);
}

int main()
{
    int n;
    cin >> n;
    vector<long long> a(n);
    for(int i=0; i<n; ++i)
        cin >> a[i];

    vector<pair<int, int> > ans;
    solve(a, ans);
    cout << ans.size() << endl;
    for(const auto& p : ans)
        cout << (p.first + 1) << ' ' << (p.second + 1) << endl;

    return 0;
}