Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

int main(){
    int n;
    cin>>n;
    
    int max=0;//絶対値最大のやつ
    int maxindex=0;//それの番号
    
    for(int i=0;i<n;i++){
        int a;
        cin>>a;
        if(abs(a)>=abs(max)){
            max=a;
            maxindex=i;
        }
    }
    
    maxindex++;//1-indexedに変換
    
    int m=2*n-1;//操作の回数
    cout<<m<<endl;
    
    //Step 1:絶対値最大のやつを他に足して符号を揃える
    for(int i=1;i<=n;i++){
        cout<<maxindex<<" "<<i<<endl;
    }
    
    //Step 2:前or後ろから順に足していく
    if(max>=0){
        for(int i=1;i<n;i++){
            cout<<i<<" "<<i+1<<endl;
        }
    }else{
        for(int i=n;i>1;i--){
            cout<<i<<" "<<i-1<<endl;
        }
    }
    
    return 0;
}