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 <algorithm>
#include <bitset>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>
#define ll long long
#define imfor(i, n) for(int i = 0; i < n; i++)
using namespace std;
ll MOD = 1000000007;

int main() {
    int n;
    int a[50];
    cin >> n;
    int absmax = 0;
    int absmaxi;
    imfor(i, n) {
        cin >> a[i];
        if (abs(a[i]) > absmax) {
            absmax = a[i];
            absmaxi = i;
        }
    }
    cout << 2 * n - 1 << endl;
    imfor (i, n) {
        a[i] += absmax;
        cout << absmaxi + 1 << " " << i + 1 << endl;
    }
    if (absmax > 0) {
        imfor (i, n - 1) {
            cout << i + 1 << " " << i + 2 << endl;
        }
    }
    else {
        for (int i = n; i > 1; i--) {
            cout << i << " " << i - 1 << endl;
        }
    }
}