Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
-2 5 -1

Sample Output 1

2
2 3
3 3

After the first operation, a = (-2,5,4).

After the second operation, a = (-2,5,8), and the condition is now satisfied.

Sample Input 2

2
-1 -3

Sample Output 2

1
2 1

After the first operation, a = (-4,-3) and the condition is now satisfied.

Sample Input 3

5
0 0 0 0 0

Sample Output 3

0

The condition is satisfied already in the beginning.

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

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

int n;
int a[60];
int Max = 0;

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

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

   setMax();

   // 全要素に足す
   int ans[2][60];
   int cnt = 0;
   for (int i = 0; i < n; i ++) {
      ans[0][cnt] = Max+1, ans[1][cnt] = i+1;
      cnt++;
      a[i] += a[Max];
   }

   // 累積和
   if ( a[Max] >= 0 ) {
      for (int i = 1; i < n; i++) {
         a[i] += a[i] + a[i-1];
         ans[0][cnt] = i, ans[1][cnt] = i+1;
         cnt++;
      }
   } else {
      for (int i = n-2; i >= 0; i--) {
         a[i] += a[i] + a[i+1];
         ans[0][cnt] = i+2, ans[1][cnt] = i+1;
         cnt++;
      }
   }

   cout << cnt << endl;
   for (int i = 0; i < cnt; i++) {
      cout << ans[0][i] << " " << ans[1][i] << endl;
   }

   // for (int i = 0; i < n; i++) {
   //    cout << a[i] << " ";
   // }
   // cout << endl;
     
   return 0;
}