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

Solution (in CPP):

Solution:
#include <cstdio>
#include <iostream>
#include <queue>
using namespace std;

int main(){
  int n;
  cin >> n;
  int ima; int imi;
  int ma = 0; int mi = 0;
  for (int i = 1; i <= n; i++){
    int k;
    scanf("%d", &k);
    if (ma <= k) {
      ima = i;
      ma = k;
    }
    if (mi >= k) {
      imi = i;
      mi = k;
    }
  }
  if (imi == ima) {
    printf("0\n");
    return 0;
  }
  printf("%d\n", 2*n-1);
  if (ma + mi >= 0) {
    for (int i = 1; i <= n; i++){
      printf("%d %d\n", ima, i);
    }
    for (int i = 1; i < n; i++){
      printf("%d %d\n", i, i+1);
    }
  } else {
    for (int i = 1; i <= n; i++){
      printf("%d %d\n", imi, i);
    }
    for (int i = n; i > 1; i--){
      printf("%d %d\n", i, i-1);
    }
  }
  return 0;
}