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> //C++の標準ライブラリを一行で一括でインクルードする
#include <math.h> //数学関数と数学定数を利用する
#define rep(i,n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;
template<class T> void chmax(T &a,T b) { if (a<b) a=b;}
template<class T> void chmin(T &a,T b) { if (a>b) a=b;}

const int INF = 1 << 30;
int main(){
  int n; cin >> n;
  vector<int> a(n);
  rep(i,n) cin >> a[i];
  pair<int,int> mi={INF,0},ma={-INF,0};
  
  rep(i,n){
    if(a[i] < mi.first) mi = {a[i],i};
    if(a[i] > ma.first) ma = {a[i],i};
  }
  
  pair<int,int> p; bool reverse = false;
  if(abs(mi.first) > abs(ma.first)){
    p = mi; reverse = true;
  }else p = ma;
  
  printf("%d\n",n+n-1);
  
  rep(i,n){
    printf("%d %d\n",p.second+1,i+1);
  }
  if(reverse){
    for(int i = n; i >= 2;i--){
      printf("%d %d\n",i,i-1);
    }
  }else{
    rep(i,n-1){
      printf("%d %d\n",i+1,i+2);
    }
  }
  
  return 0;
}