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;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define INF 1<<30
#define LINF 1LL<<60
vector<pii> ans;
vector<ll> a;
ll N;
bool dfs(int n,int dep) {
	if (n == N - 1) return true;
	if (dep == 2 * N) {
		bool f = true;
		for (int i = 0; i < N - 1;i++) {
			if (a[i] > a[i + 1]) {
				f = false;
				break;
			}
		}
		return f;
	}
	while (a[n] > a[n + 1]) {
		n++;
		if (n == N) return true;
	}
	bool f;
	if (a[n] >= 0 && a[n + 1] >= 0) {
		if (a[n] != 0) {
			a[n + 1] += a[n];
			f = dfs(n, dep + 1);
			a[n + 1] -= a[n];
			if (f) {
				ans.push_back({ n,n + 1 });
				return true;
			}
		}
		if (a[n + 1] != 0) {
			a[n + 1] += a[n + 1];
			f = dfs(n, dep + 1);
			a[n + 1] -= a[n + 1];
			if (f) {
				ans.push_back({ n + 1,n + 1 });
				return true;
			}
		}
	}else if (a[n] > 0 && a[n + 1] < 0) {
		a[n + 1] += a[n];
		f = dfs(n, dep + 1);
		a[n + 1] -= a[n];
		if (f) {
			ans.push_back({ n,n+1 });
			return true;
		}
	} if (a[n] < 0 && a[n + 1] < 0) {
		a[n] += a[n];
		f = dfs(n, dep + 1);
		a[n] -= a[n];
		if (f) {
			ans.push_back({ n,n });
			return true;
		}

		a[n] += a[n + 1];
		f = dfs(n, dep + 1);
		a[n] -= a[n + 1];
		if (f) {
			ans.push_back({ n + 1 , n});
			return true;
		}
	}
	return false;
}
int main(void) {
	cin.tie(0); ios::sync_with_stdio(false);
	cin >> N;
	a.resize(N);
	for (auto &in : a) cin >> in;
	dfs(0, 0);
	cout << ans.size() << endl;
	for (auto p : ans) {
		cout << p.first + 1 << " " << p.second + 1 << endl;
	}
	return 0;
}