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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mp make_pair
#define pb push_back
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)

int main(){
  int N;
  cin >> N;
  int a[N];
  rep(i,N) cin >> a[i];
  int mmax=0;
  int pmax=0;
  int mind, pind;
  rep(i,N){
    if(a[i]<0){
      if(mmax > a[i]){
	mmax=a[i];
	mind=i;
      }
    }
    else if(a[i]>0){
      if(pmax < a[i]){
	pmax=a[i];
	pind=i;
      }
    }    
  }
  rep1(i,N-1) {
    if(a[i-1] <= a[i]){
      //      cout << i;
      if(i==N-1){
	cout << 0 << endl;
	return 0;
      }
    }
    else {
      break;
    }
  }
  if(mmax==0 && pmax==0){
    cout << 0 << endl;
    return 0;
  }
  cout << mmax << " " << pmax << endl;
  vector<pair<int, int> > ans;  
  if(abs(mmax) < abs(pmax)){
    rep(i,N){
      if(a[i] < 0){
	ans.pb(mp(pind,i));
	a[i]+=a[pind];
      }
    }
    rep1(i,N-1){
      if(a[i-1] > a[i] ){
	ans.pb(mp(i-1,i));
	a[i]+=a[i-1];
      }
    }
  }
  else{
    rep(i,N){
      if(a[i] > 0){
        ans.pb(mp(mind,i));
	a[i]+=a[mind];
      }
    }
    for(int i=N-1; i>0; i--){
      if(a[i-1] > a[i] ){
	ans.pb(mp(i,i-1));
	a[i-1]+=a[i];
      }
    }
  }
  cout << ans.size() << endl;  
  rep(i,ans.size()){
    cout << ans[i].first + 1 << " " << ans[i].second + 1 << " " << endl;
  }
  
  return 0;

}