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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <utility>
using namespace std;
#define ll long long

pair<ll, ll> box[100];
ll a[50], sub[50];
ll n;
bool ok = false;

void dfs (ll itr) {
  if (ok) {
    return ;  
  }
  if (itr == 2 * n) {
    for (ll p = 0; p < 2 * n; p++) {
      ll i = box[p].first;
      ll j = box[p].second;
      sub[j] = sub[j] + sub[i];    
    }  
    ll now = sub[0];
    for (ll p = 1; p < n; p++) {
      if (now <= sub[p]) {
        now = sub[p];
      }    
      else {
        for (ll q = 0; q < n; q++) {
          sub[q] = a[q];  
        }
        return ;    
      }
    }
    cout << 2 * n << endl;
    for (ll p = 0; p < 2 * n; p++) {
      cout << box[p].first + 1 << " " << box[p].second + 1<< endl;  
    }
    ok = true;
  }  
  else {
    for (ll i = 0; i < n; i++) {
      for (ll j = 0; j < n; j++) {
        if (i != j) {
          box[itr] = make_pair(i, j);
          dfs(itr + 1); 
        }
      }  
    }  
  }
}



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