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>

using namespace std;

int main(){
    int N;
    int a[50];
    cin>>N;
    int AbsMaxInd;
    int AbsMax=-1;
    for(int i=0;i<N;i++){
        cin>>a[i];
        if(abs(a[i]) > AbsMax){
            AbsMax=abs(a[i]);
            AbsMaxInd=i;
        }
    }
    AbsMaxInd++;
    if(AbsMax==0){
        cout<<0<<endl;
    }else if(a[AbsMaxInd-1]>0){//絶対値の一番大きい数が正の時
        cout<<(2*(N-1))<<endl;
        for(int i=1;i<=N;i++){
            if(i != AbsMaxInd){
                cout<< AbsMaxInd << ' ' << i <<endl;
            }
        }
        for(int i=1;i<N;i++){
            cout<< i << ' ' << (i+1) <<endl;
        }
    }else{//絶対値の一番大きい数が負の時
        cout<<(2*(N-1))<<endl;
        for(int i=1;i<=N;i++){
            if(i != AbsMaxInd){
                cout<< AbsMaxInd << ' ' << i <<endl;
            }
        }
        for(int i=N;i>1;i--){
            cout<< i << ' ' << (i-1) <<endl;
        }
    }
    
    
    return 0;
}
