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<vector>
#include<algorithm>
#include<stdlib.h>
using namespace std;

int main() {
    int n;
    cin>>n;
    vector<int> a(n);

    for(int i=0; i<n; i++) {
        cin>>a[i];
    }

    vector<int>::iterator maxItr = max_element(a.begin(), a.end());
    vector<int>::iterator minItr = min_element(a.begin(), a.end());
    size_t maxIndex = distance(a.begin(), maxItr);
    size_t minIndex = distance(a.begin(), minItr);
    cout<<2*n-2<<endl;
    if(abs(*maxItr)>abs(*minItr)) {
        for(int i=0; i<n; i++) {
            if(i==maxIndex) continue;
            cout<<maxIndex+1<<" "<<i+1<<endl;
        }
        for(int i=0; i<n-1; i++) {
            cout<<i+1<<" "<<i+2<<endl;
        }
    } else{
        for(int i=0; i<n; i++) {
            if(i==minIndex) continue;
            cout<<minIndex+1<<" "<<i+1<<endl;
        }
        for(int i=n-1; i>0; i--) {
            cout<<i+1<<" "<<i<<endl;
        }
    }
}