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;

using ll = long long;
#define debug(x) cerr << #x << ": " << x << endl;
#define line() cerr << "---------------" << endl;

int a[60];

int main() {
   int n;
   cin >> n;
   for (int i = 0; i < n; i++) {
      cin >> a[i];
   }
   int Max = 0;
   for (int i = 1; i < n; i++) {
      if ( abs(a[i]) > abs(a[Max]) ) {
         Max = i;
      }
   }

   // 全要素に足す
   cout << 2 * n - 1 << endl;
   int cnt = 0;
   for (int i = 0; i < n; i++) {
      a[i] += a[Max];
      cout << Max + 1 << " " << i + 1 << endl;
      cnt++;
   }
   
   // 累積和
   if (a[Max] >= 0) {
      for (int i = 1; i < n; i++) {
         a[i] += a[i] + a[i-1];
         cout << i << " " << i + 1 << endl;
         cnt++;
      }
   } else {
      for (int i = n-2; i >= 0; i--) {
         a[i] += a[i] + a[i+1];
         cout << i + 2 << " " << i + 1 << endl;
         cnt++;
      }
   }

   return 0;
}