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;
typedef long long ll;
typedef pair<int,int> pii;
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
#define all(a) (a).begin(),(a).end()
#define pb emplace_back
#define INF (1e9+1)

int main(){
    int n;
    cin>>n;
    
    vector<int> v(n);
    rep(i,n)cin>>v[i];
    
    int mi = *min_element(all(v));
    int mx = *max_element(all(v));

    vector<pii> ans;
    if(abs(mi)>abs(mx)){
        int p = min_element(all(v)) - v.begin();
        rep(i,n){
            if(i!=p)ans.pb(pii(i,p));
        }
        for(int i=n-2;i>=0;i--){
            ans.pb(pii(i,i+1));
        }
    }else{
        int p = max_element(all(v)) - v.begin();
        rep(i,n){
            if(i!=p)ans.pb(pii(i,p));
        }
        for(int i=1;i<n;i++){
            ans.pb(pii(i,i-1));
        }
    }
    
    cout<<ans.size()<<endl;
    rep(i,ans.size()){
        cout<<ans[i].second+1<<" "<<ans[i].first+1<<endl;
    }
}
