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;
#define rep(i, N) for (int i = 0; i < (int)N; i++)
#define all(a) (a).begin(), (a).end()

int N;
vector<int> a;

struct op {
  int x, y;
};

int main () {
  cin >> N;
  a.resize(N);
  rep(i,N) cin >> a[i];

  vector<op> ans;

  int iMax = distance(a.begin(), max_element(all(a)));
  int iMin = distance(a.begin(), min_element(all(a)));

  bool right = (iMin >= 0);
  if(abs(a[iMax]) > abs(a[iMin])) {
    rep(i,N) if(a[i] < 0) {
      ans.push_back(op{iMax, i});
    }
    right = true;
  } else {
    rep(i,N) if(a[i] > 0) {
      ans.push_back(op{iMin, i});
    }
    right = false;
  }

  rep(i,N-1) {
    if(right) {
      ans.push_back(op{i, i+1});
    } else {
      ans.push_back(op{N-1-i, N-2-i});
    }
  }

  cout << ans.size() << endl;
  for(auto &e: ans) {
    cout << (e.x+1) << " " << (e.y+1) << endl;
  }

  return 0;
}
