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>

using namespace std;
typedef long long ll;

const ll INF = 1e9;
const ll MOD = 1e9 + 7;

int main() {
  int N;
  cin >> N;
  ll max_idx = -1;
  ll maxv = -1;
  vector<ll> list(N);
  for (int i = 0; i < N; i++) {
    cin >> list[i];
    if (maxv < abs(list[i])) {
      maxv = abs(list[i]);
      max_idx = i;
    }
  }

  vector<pair<ll, ll> > hist;
  for (int i = 0; i < N; i++) {
    list[i] += list[max_idx];
    hist.push_back(make_pair(max_idx + 1, i + 1));
  }

  if (list[max_idx] >= 0) {
    for (int i = 0; i < N - 1; i++) {
      list[i + 1] += list[i];
      hist.push_back(make_pair(i + 1, i + 2));
    }
  } else {
    for (int i = N - 1; i > 0; i--) {
      list[i - 1] += list[i];
      hist.push_back(make_pair(i + 1, i));
    }
  }

  cout << hist.size() << endl;
  for (auto itr = hist.begin(); itr != hist.end(); itr++) {
    cout << itr->first << " " << itr->second << endl;
  }
  return 0;
}
