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
//------------------------------------------
#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>

using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

//math
//-------------------------------------------
template<class T> inline T sqr(T x) {return x*x;}

//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;

//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
 
//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI  = acos(-1.0);

//clear memory
#define CLR(a) memset((a), 0 ,sizeof(a))

//debug
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;


int main(){
    int N,tar;
    tar = 0;
    cin >> N;
    int Nind = N-1;
    vector< pair<int,int> > seq(N, make_pair(0,0) );
    int count = 0;
    EACH(iter,seq){
        int val;
        cin >> val;
        *iter = make_pair(val, count);
        count++;
    }
   int delta;

    queue< pair<int, int> > q;
    vector< pair<int,int> >  ord = seq;
    while(tar != N){
        delta = seq[tar+1].first - seq[tar].first;
        
        if( delta >= 0 ){
            tar++;            
            continue;
        }else{
            ord = seq;
            SORT(ord);
            auto iter = lower_bound( ALL(ord), make_pair(-delta,-1) );
            if( iter == ord.end() ) iter--; 
            seq[tar+1].first += iter->first ;                 
            q.push( make_pair( iter->second + 1  ,(tar+1)+1 ) ) ;
                }
    }
    
    cout << q.size() << endl;
    while( !q.empty() ){
        auto f = q.front().first;
        auto s = q.front().second;
        cout << f << " " << s << endl;
        q.pop();
    }
}