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;
vector<pair<ll, ll> > vec;

int main() {
  ll N;
  vector<ll> a;
  ll max_abs = -1, idx;

  cin >> N;
  for (ll i = 0; i < N; i++) {
    ll t;
    cin >> t;
    a.push_back(t);
    if(abs(t) > max_abs) {
      max_abs = abs(t);
      idx = i;
    }
  }

  bool plus = (a[idx] >= 0 ? true : false);

  for (ll i = 0; i < N; i++) {
    a[i] += a[idx];
    vec.push_back({idx + 1, i + 1});
  }

  if (plus) {
    for (ll i = 0; i < N - 1; i++) {
      a[i + 1] += a[i];
      vec.push_back({i + 1, i + 2});
    }
  } else {
    for (ll i = N -1;  i >= 1; i--) {
      a[i - 1] += a[i];
      vec.push_back({i + 1, i});
    }
  }

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