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>
using namespace std;

int main(){
    int N;
    cin >> N;

    long long a[50];
    long long max=-1000001,min=1000001;
    int imax=0,imin=0;

    for(int i=0;i<N;i++){
        cin >> a[i];
        if(a[i]>max){ max=a[i]; imax=i+1; }
        if(a[i]<min){ min=a[i]; imin=i+1; }
    }

    if(min>=0){
        cout << (N-1) << endl;
        for(int i=1;i<=N-1;i++) cout << i << ' ' << (i+1) << endl;
    }else if(max<=0){
        cout << (N-1) << endl;
        for(int i=N-1;i>=1;i--) cout << (i+1) << ' ' << i << endl;
    }else if(max>=-min){
        cout << (2*N-1) << endl;
        for(int i=1;i<=N;i++) cout << imax << ' ' << i << endl;
        for(int i=1;i<=N-1;i++) cout << i << ' ' << (i+1) << endl;        
    }else{
        cout << (2*N-1) << endl;
        for(int i=1;i<=N;i++) cout << imin << ' ' << i << endl;
        for(int i=N-1;i>=1;i--) cout << (i+1) << ' ' << i << endl;
    }

    return 0;
}