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>
using namespace std;
/* typedef */
typedef long long ll;
typedef pair<int, int> pii;
/* constant */
const int INF = 1 << 30;
const ll LINF = 1LL << 61;
const int NIL = -1;
const int MAX = 10000;
const int mod = 1000000007;
const double pi = 3.141592653589;
/* global variables */
/* function */
/* main */
int main(){
    int n;
    cin >> n;
    vector<int> A(n);
    for (int &a : A) {
        cin >> a;
    }
    // MAX := max(a), MIN := min(a)
    // |MAX| ≥ |MIN| -> すべての数に MAX を足せば ai > 0とできる
    // |MAX| < |MIN| -> すべての数に MIN を足せば ai <= 0とできる
    // あとは累積和
    int mx_ind = max_element(A.begin(), A.end()) - A.begin();
    int mi_ind = min_element(A.begin(), A.end()) - A.begin();
    int mx = A[mx_ind], mi = A[mi_ind];
    vector<pii> ans;
    if (abs(mx) >= abs(mi)) {
        // a[i] += mx
        for (int i = 0; i < n; i++) ans.push_back(pii(mx_ind, i));
        // a[i + 1] += a[i]
        for (int i = 0; i < n - 1; i++) ans.push_back(pii(i, i + 1));
    }
    else {
        // a[i] += mi
        for (int i = 0; i < n; i++) ans.push_back(pii(mi_ind, i));
        // a[i - 1] += a[i]
        for (int i = n - 1; i > 0; i--) ans.push_back(pii(i, i - 1));
    }

    cout << ans.size() << '\n';
    for (pii p : ans) {
        cout << p.first + 1 << ' ' << p.second + 1<< '\n';
    }

}