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 <bits/stdc++.h>

using namespace std;
using LL = long long;
const LL LINF = 1e18;
const LL MOD = 1e9+7;


int main(){
    int N;cin >> N;
    vector<int> vec;
    vector<pair<int,int>> ans;
    for(int a = 0;a < N;a++){
        int b;cin >> b;
        vec.push_back(b);
    }
    if(abs(*max_element(vec.begin(),vec.end())) > abs(*min_element(vec.begin(),vec.end()))){
        for(int a = 0;a < N;a++){
            ans.push_back({max_element(vec.begin(),vec.end())-vec.begin(), a});
        }
        for(int a = 0;a < N-1;a++){
            ans.push_back({a,a+1});
        }
    }else{
        for(int a = 0;a < N;a++){
            ans.push_back({min_element(vec.begin(),vec.end())-vec.begin(), a});
        }
        for(int a = N-1;a >= 1;a--){
            ans.push_back({a,a-1});
        }
    }
    cout<<ans.size()<<endl;
    for(int a = 0;a < ans.size();a++){
        cout<<ans.at(a).first+1<<" "<<ans.at(a).second+1<<endl;
    }
}
