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;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb push_back
#define mp make_pair
typedef long long ll;
typedef pair<int,int> pint;

int a[50],b[50];
vector<pint> pi;
int main(){
    int n;
    cin>>n;
    rep(i,n) cin>>a[i];
    int mx=max_element(a,a+n)-a;
    int mn=min_element(a,a+n)-a;
    bool flag=true;
    if(abs(a[mx])<abs(a[mn])) flag=false;
    if(flag){
        rep(i,n)if(i!=mx){
            a[i]+=a[mx];
            pi.pb(mp(mx+1,i+1));
        }
        rep(i,n-1) pi.pb(mp(i+1,i+2));
    } 
    else{
        rep(i,n)if(i!=mn){
            a[i]+=a[mn];
            pi.pb(mp(mn+1,i+1));
        }
        for(int i=n-1;i>=1;--i) pi.pb(mp(i+1,i));
    }
    cout<<pi.size()<<endl;
    rep(i,pi.size()) cout<<pi[i].first<<" "<<pi[i].second<<endl;
    return 0;
}