#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair <int,int> P;

int main(){
  int n;
  cin>>n;
  cout<<2*n-1<<endl;
  int a[55];
  int ma=0,ma_i=0;
  rep(i,n){
    cin>>a[i];
    if(abs(a[i])>abs(ma)){
      ma=a[i];
      ma_i=i;
    }
  }
  rep(i,n){
    a[i]+=ma;
    cout<<ma_i+1<<' '<<i+1<<endl;
  }
  if(ma>=0)
    rep(i,n){
      if(i>0){
	a[i]+=a[i-1];
	cout<<i<<' '<<i+1<<endl;
      }
    }
  else
    for(int i=n-1;i>=0;i--){
      if(i<n-1){
	a[i]+=a[i+1];
	cout<<i+2<<' '<<i+1<<endl;
      }
    }
  return 0;
}



### Prompt

Please create a solution in Cpp to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair <int,int> P;

int main(){
  int n;
  cin>>n;
  cout<<2*n-1<<endl;
  int a[55];
  int ma=0,ma_i=0;
  rep(i,n){
    cin>>a[i];
    if(abs(a[i])>abs(ma)){
      ma=a[i];
      ma_i=i;
    }
  }
  rep(i,n){
    a[i]+=ma;
    cout<<ma_i+1<<' '<<i+1<<endl;
  }
  if(ma>=0)
    rep(i,n){
      if(i>0){
	a[i]+=a[i-1];
	cout<<i<<' '<<i+1<<endl;
      }
    }
  else
    for(int i=n-1;i>=0;i--){
      if(i<n-1){
	a[i]+=a[i+1];
	cout<<i+2<<' '<<i+1<<endl;
      }
    }
  return 0;
}



```