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.

Examples

Input

3
-2 5 -1


Output

2
2 3
3 3


Input

2
-1 -3


Output

1
2 1


Input

5
0 0 0 0 0


Output

0
#include <bits/stdc++.h>
#define _CRT_SECURE_NO_WARNINGS
#define ll long long
#define BUF 50
#define INF 1<<30
using namespace std;
ll MOD = 1e9 + 7;
ll A, B, C, D, G, H, N, M, L, K, P, Q, R, W, X, Y, Z;
string S;
ll ans = 0;
int main() {
	cin >> N;
	vector<int>a(N);
	vector<pair<int, int>>res;
	for (int i = 0; i < N; i++)cin >> a[i];
	auto mxit = max_element(a.begin(), a.end());
	auto mnit = min_element(a.begin(), a.end());
	if (*mnit < 0 && *mxit < abs(*mnit)) {
		for (int i = 1; i <= N; i++)res.push_back({ distance(a.begin(),mnit) + 1, i });
		for (int i = N; i > 1; i--)res.push_back({ i, i - 1 });
	}
	else {
		for (int i = 1; i <= N; i++)res.push_back({ distance(a.begin(),mxit) + 1,i });
		for (int i = 1; i < N; i++)res.push_back({ i,i + 1 });
	}
	cout << res.size() << endl;
	for (int i = 0; i < res.size(); i++)cout << res[i].first << " " << res[i].second << endl;
}