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 <bits/stdc++.h>

using namespace std;

int main()
{
	int N;
	cin >> N;
	int a[55] = {};
	for (int i = 0; i < N; i++)
		cin >> a[i + 1];
	vector<pair<int, pair<int, int>>> b;
	for (int i = 0; i < N + 1; i++)
		for (int j = 0; j < N + 1; j++)
			b.push_back(make_pair(a[i] + a[j] + (j == 1 ? a[i] : 0), make_pair(i, j)));
	sort(b.begin(), b.end());
	vector<pair<int, int>> r;
	if (b[0].second.first) {
		r.push_back(make_pair(b[0].second.first, 1));
		r.push_back(make_pair(b[0].second.first, 1));
	}
	a[1] += b[0].first;
	for (int i = 2; i < N + 1; i++) {
		b.clear();
		for (int k = 0; k < N + 1; k++)
			for (int j = 0; j < N + 1; j++)
				b.push_back(make_pair(a[k] + a[j] + (j == i ? a[k] : 0), make_pair(k, j)));
		sort(b.begin(), b.end());
		auto c = make_pair(a[i - 1] - a[i], make_pair(0, 0));
		auto l = lower_bound(b.begin(), b.end(), c);
		if (l->second.first)
			r.push_back(make_pair(l->second.first, i));
		if (l->second.second)
			r.push_back(make_pair(l->second.second, i));
		a[i] += l->first;
	}
	cout << r.size() << '\n';
	for (auto &p : r)
		cout << p.first << ' ' << p.second << '\n';
	cout << flush;
	return 0;
}