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> P;
#define rep(i,m,n) for(int i=m;i<n;i++)
ll mod=1e9+7;

int main(){
  int n;
  cin>>n;
  int a[n];
  rep(i,0,n) cin>>a[i];
  P b[n];
  rep(i,0,n) b[i]=P(a[i],i);
  sort(b,b+n);
  if(b[0].first>=0){
    cout<<n-1<<endl;
    rep(i,0,n-1){
      cout<<i+1<<' '<<i+2<<endl;
      a[i+1]+=a[i];
    }
  }
  else if(b[0].first+b[n-1].first>=0){
    cout<<2*n-2<<endl;
    rep(i,0,n){
      if(i==b[n-1].second) continue;
      cout<<b[n-1].second+1<<' '<<i+1<<endl;
      a[i]+=b[n-1].first;
    }
    rep(i,0,n-1){
      cout<<i+1<<' '<<i+2<<endl;
      a[i+1]+=a[i];
    }
  }
  else{
    cout<<2*n-2<<endl;
    rep(i,0,n){
      if(i==b[0].second) continue;
      cout<<b[0].second+1<<' '<<i+1<<endl;
      a[i]+=b[0].first;
    }
    rep(i,0,n-1){
      cout<<n-i<<' '<<n-1-i<<endl;
      a[n-2-i]+=a[n-1-i];
    }
  }
}