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 <cstdio>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <algorithm>
#include <utility>

int main() {
  size_t n;
  scanf("%zu", &n);

  std::vector<int> a(n);
  for (auto& ai: a) scanf("%d", &ai);

  size_t k = 0;
  for (size_t i = 1; i < n; ++i)
    if (std::abs(a[k]) < std::abs(a[i]))
      k = i;

  if (a[k] == 0) return puts("0"), 0;

  std::vector<std::pair<size_t, size_t>> res;
  for (size_t i = 0; i < n; ++i) {
    if (i != k) res.emplace_back(k, i);
  }

  if (a[k] > 0) {
    for (size_t i = 1; i < n; ++i)
      res.emplace_back(i-1, i);
  } else {
    for (size_t i = n-1; i--;)
      res.emplace_back(i+1, i);
  }

  printf("%zu\n", res.size());
  for (auto const& p: res)
    printf("%zu %zu\n", p.first+1, p.second+1);

  {
    for (auto const& p: res)
      a[p.second] += a[p.first];

    assert(std::is_sorted(a.begin(), a.end()));
  }
}
