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>
using namespace std;
using ll = long long;

int main() {
	int N;
	cin >> N;
	vector<int> A(N);
	for(auto &a : A) cin >> a;
	int minv = INT_MAX, maxv = INT_MIN;
	int minp = -1, maxp = -1;
	for(int i=0;i<N;i++) {
		if(minv > A[i]) {
			minp = i;
			minv = A[i];
		}
		if(maxv < A[i]) {
			maxp = i;
			maxv = A[i];
		}
	}
	vector<pair<int,int>> ans;
	if(abs(minv) < abs(maxv)) {
		for(int i=0;i<N;i++) {
			if(A[i] < 0) {
				ans.push_back({maxp ,i});
				A[i] += maxv;
			}
		}
		for(int i=1;i<N;i++) {
			A[i] += A[i-1];
			ans.push_back({i-1, i});
		}
	} else {
		for(int i=0;i<N;i++) {
			if(A[i] > 0) {
				ans.push_back({minp ,i});
				A[i] += minp;
			}
		}
		for(int i=N-2;i>=0;i--) {
			A[i] += A[i+1];
			ans.push_back({i+1, i});
		}
	}
	cout << ans.size() << endl;
	for(auto &a : ans) cout << a.first+1 << " " << a.second+1 << endl;

	return 0;
}