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
//競技プログラミング　テンプレ
//負の時は逆!!
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <map>
#include <cstdlib>

using namespace std;

#define REP(i,n) for(int i = 0; i < (int)(n); ++i)

int main() {
	int n, ans=0,max_n,min_n;
	bool f=0;
	cin >> n;
	vector <int> A(n);
	//string str;
	for (int i = 0; i < n; i++) {
		cin >> A[i];
	}
	int min = 1000001, max = -1000001;
	REP(i,n){
		if(max < A[i]) {max = A[i]; max_n  =i;}
		if(min > A[i]) {min = A[i]; min_n = i;}
	}
	//sort(A.begin(), A.end());
	if( abs(A[min_n]) < abs(A[max_n]) ) f=1;
	//cout << f << endl;

	cout << 2*n-2 << endl;
	if(f){
		REP(i,n) { if( (i+1) != max_n + 1) cout << max_n+1 << " "<<  i+1 << endl; }
		REP(i,n-1) cout << i+1 <<" "<< i+2 << endl;
	}
	else{
		REP(i,n) { if(min_n + 1 != (i+1))cout << min_n+1 << " " << i+1 << endl; }
		REP(i,n-1) cout << n-i <<" "<< n-i-1 << endl;
	}


	return 0;
}
