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 <iostream>
#include <algorithm>
#include <vector>
#include <string.h>
typedef long long ll;

#define MAX(x, y)	(((x) > (y))? (x) : (y))
#define MIN(x, y)	(((x) < (y))? (x) : (y))
#define ABS(x)		(((x) < 0)? (-1*(x)) : (x))
#define REPI(x,y)	for (int i=(x);i<(y);i++)
#define REPJ(x,y)	for (int j=(x);j<(y);j++)
#define MEMCLR(X)	memset((X),0,sizeof((X)))

using namespace std;
typedef vector<vector<pair<int, int>>> GRA;

int i,j,N,K,ma,mi=1000000,mai,mii;

int main() {
	cin >> N;
	
	vector<int> A(N);

	REPI(0,N) {
		cin >> A[i];
		if (A[i]>ma) {
			ma = A[i];
			mai = i;
		} else if (A[i]<mi) {
			mi = A[i];
			mii = i;
		}
	}

	cout << (2*N-1) << endl;
	
	REPI(1,N+1) {
		if (ABS(ma)>=ABS(mi)) cout << (mai+1) << " " << i << endl;
		else cout << (mii+1) << " " << i << endl;
		
	}

	REPI(0,N-1) {
		if (ABS(ma)>=ABS(mi)) cout << (i+1) << " " << i+2 << endl;
		else cout << (N-i) << " " << (N-i-1) << endl;
	}
	
	return 0;
}