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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <fstream>
#include <vector>
#include <set>
#include <algorithm>
#include <queue>
#include <map>
#include <unordered_map>
#include <random>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;
using ll = long long;

ll N;
vector<ll> A;

void input() {
    cin >> N;
    A.resize(N);
    for (auto& a : A)
        cin >> a;
}

void solve() {
    auto maxi = max_element(A.begin(), A.end());
    auto mini = min_element(A.begin(), A.end());
    auto put_ans = [](ll l, ll r) {
        cout << l + 1 << ' ' << r + 1 << endl;
    };
    if (abs(*maxi) >= abs(*mini)) {
        for (auto it = A.begin(); it != A.end(); ++it) {
            if (it != maxi)
                put_ans(maxi - A.begin(), it - A.begin());
        }
        for (ll i = 0; i < N - 1; i++)
            put_ans(i, i + 1);
    } else {
        for (auto it = A.begin(); it != A.end(); ++it) {
            if (it != mini)
                put_ans(mini - A.begin(), it - A.begin());
        }
        for (ll i = N - 1; i > 0; i--)
            put_ans(i, i - 1);
    }
}


int main() {
    input();
    solve();
    
    return 0;
}