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>
typedef long long ll;
using namespace std;

ll n, a[51];
int main(){
  cin >> n;
  ll mini = INT_MAX, maxi = INT_MIN;
  int argmini, argmaxi;
  for(int i=1;i<=n;i++){
    cin >> a[i];
    if (mini > a[i]) {mini = a[i]; argmini = i;}
    if (maxi < a[i]) {maxi = a[i]; argmaxi = i;}
  }
  if (0 <= mini){
    cout << n-1 << endl;
    for(int i=1;i<n;i++) cout << i << " " << i+1 << endl;
  }else if (maxi <= 0){
    cout << n-1 << endl;
    for(int i=n;i>1;i--) cout << i << " " << i-1 << endl;    
  }else{
    if (abs(mini) <= maxi){
      cout << 2*n-1 << endl;
      for(int i=1;i<=n;i++) cout << argmaxi << " " << i << endl;
      for(int i=1;i<n;i++) cout << i << " " << i+1 << endl;
    }else{
      cout << 2*n-1 << endl;
      for(int i=1;i<=n;i++) cout << argmini << " " << i << endl;
      for(int i=n;i>1;i--) cout << i << " " << i-1 << endl;
    }
  }
}
