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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <iostream>

using namespace std;

typedef long long ll;

int main() {
    ll N, A[51];

    cin >> N;
    ll maxAbs = 0;
    int maxAbsIdx = 0;

    for (int i = 1; i <= N; i++) {
        cin >> A[i];
        if (abs(A[i]) > maxAbs) {
            maxAbs = abs(A[i]);
            maxAbsIdx = i;
        }
    }

    if (maxAbs == 0) {
        cout << 0 << endl;
        return 0;
    }

    vector<string> ans;
    
    for (int i = 1; i <= N; i++) {
        if (i == maxAbsIdx) { continue; }
        A[i] += A[maxAbsIdx];
        char s[100];
        sprintf(s, "%d %d\n", maxAbsIdx, i);
        ans.push_back(s);
    }

    if (A[maxAbsIdx] > 0) {
        for (int i = 2; i <= N; i++) {
            A[i] += A[i-1];
            char s[100];
            sprintf(s, "%d %d\n", i-1, i);
            ans.push_back(s);
        }
    } else {
        for (int i = N - 1; i >= 1; i--) {
            A[i] += A[i+1];
            char s[100];
            sprintf(s, "%d %d\n", i+1, i);
            ans.push_back(s);
        }
    }

    cout << ans.size() << endl;
    for (const auto& e : ans) {
        std::cout << e;
    }
}