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>

using namespace std;

typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<bool> vb;
typedef vector<string> vst;
typedef pair<ll,ll> pll;
typedef vector<pll> vpll;


const ll mod=(1e+9)+7;
const ll inf=1e+16; 

#define rep(i,m,n) for(ll i=m;i<n;i++)
#define repr(i,m,n) for(ll i=m;i>=n;i--)

int main(){
   cin.tie(nullptr);
  ios::sync_with_stdio(false);
 
 ll n,m=0,index;
  cin >> n;
  vll a(n);
  vpll res;
 
  rep(i,0,n) {
    cin >> a[i];
    if(m<abs(a[i])) {
      index=i;
      m=abs(a[i]);
    }
    }
  
  rep(i,0,n) if(a[index]*a[i]<0) res.push_back(make_pair(index+1,i+1));
  
  if(a[index]>=0) rep(i,0,n-1) res.push_back(make_pair(i+1,i+2));
  else repr(i,n-1,1) res.push_back(make_pair(i+1,i));
  
  
  
  
 
 cout << res.size() << endl;
 for(auto p:res) cout << p.first << " " << p.second << endl;
  
}
