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;

int n, mnum = 0, num = 0;
int a[55] = {0};
vector<pair<int, int>> ans;

void solve();

int main() {
  cin >> n;
  for(int i = 0; i < n; ++i) cin >> a[i];
  solve();
  return 0;
}

void solve() {
  for(int i = 0; i < n; ++i)
    if(abs(mnum) < abs(a[i])) mnum = a[i], num = i;
  for(int i = 0; i < n; ++i)
    if(mnum != a[i]) {
      ans.push_back(make_pair(num, i));
      a[i] += mnum;
    }
  if(mnum > 0) {
    for(int i = 1; i < n; ++i)
      if(a[i - 1] > a[i]) {
        ans.push_back(make_pair(i - 1, i));
        a[i] += a[i - 1];
      }
  }
  else {
    for(int i = n - 1; i >= 1; --i)
      if(a[i - 1] > a[i]) {
        ans.push_back(make_pair(i, i - 1));
        a[i - 1] += a[i];
      }
  }
  cout << ans.size() << endl;
  for(int i = 0; i < ans.size(); ++i)
    cout << ans[i].first + 1 << " " << ans[i].second + 1
         << endl;
}
