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>

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()

using namespace std;

int main(){
  int N;
  cin >> N;
  vector<int> a(N);
  int minNumIndex,maxNumIndex;
  rep(i,N){
    cin >> a[i];
    if(i == 0){
      minNumIndex = i;
      maxNumIndex = i;
    }else{
      if(a[i] < a[minNumIndex])   minNumIndex = i;
      if(a[i] >= a[maxNumIndex])  maxNumIndex = i;
    }
  }
  cout << 2*N-1 << "\n";
  if(abs(a[minNumIndex]) >= abs(a[maxNumIndex])){
    rep(i,N){
      cout << minNumIndex +1<< ' ' << i +1<< "\n";
    }
    for(int i = N-1; i >= 1; i--) {
      cout << i +1<< ' ' << i-1 +1<< "\n";
    }

  }else{
    rep(i,N){
      cout << maxNumIndex +1<< ' ' << i +1<< "\n";
    }
    rep(i,N-1){
      cout << i +1<< ' ' << i+1 +1<< "\n";
    }
  }

  return 0;
}
