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 <vector>
#include <algorithm>
#include <string>
#include <chrono>
using namespace std;

class Timer {
	chrono::high_resolution_clock::time_point start, end;
	double limit;

public:
	Timer() {
		start = chrono::high_resolution_clock::now();
	}
	Timer(double l) {
		start = chrono::high_resolution_clock::now();
		limit = l;
	}

	double getTime() {
		end = chrono::high_resolution_clock::now();
		return chrono::duration<double>(end - start).count();
	}

	bool Over() {
		if (getTime() > limit) {
			return true;
		}
		return false;
	}

	void setLimit(double l) {
		limit = l;
	}
	void setStart() { start = chrono::high_resolution_clock::now(); }
	double getLimit() { return limit; }
};
class Xor128 {
	unsigned static int x, y, z, w;
public:
	Xor128() {
		x = 31103110, y = 123456789, z = 521288629, w = 88675123;
	}

	unsigned int rand()
	{
		unsigned int t;
		t = (x ^ (x << 11)); x = y; y = z; z = w;
		return(w = (w ^ (w >> 19)) ^ (t ^ (t >> 8)));
	}
};
unsigned int Xor128::x, Xor128::y, Xor128::z, Xor128::w;


int main()
{
	Timer tmr(1.9); Xor128 xor128;
	int N; cin >> N;
	vector<int>a(N);
	for (int i = 0; i < N; i++) {
		cin >> a[i];
	}

	while (!tmr.Over()) {
		vector<int>b(a);
		vector<pair<int, int>>order(2 * N);
		for (int i = 0; i < 2 * N; i++) {
			int x, y;
			x = xor128.rand() % N;
			y = xor128.rand() % N;
			order[i] = make_pair(x, y);

			b[x] += b[y];
		}

		bool flag = true;
		for (int i = 0; i < N - 1; i++)if (b[i + 1] < b[i])flag = false;

		if (flag) {
			for (int i = 0; i < 2 * N; i++)cout << order[i].first + 1 << " " << order[i].second + 1 << endl;
			return 0;
		}
	}

	cout << "ダメ" << endl;

	return 0;
}