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 <iostream>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <math.h>
#include <cassert>
#define rep(i,n) for(int i = 0; i < n; ++i )
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  int ma = -1e9, mi = 1e9;
  int mai,mii;
  rep(i,n){
    cin >> a[i];
    ma = max(ma,a[i]);
    mi = min(mi,a[i]);
    if(ma==a[i]) mai=i;
    if(mi==a[i]) mii=i;
  }
  ++mai; ++mii;

  vector<P> ans;
  if(abs(ma)>=abs(mi)){
    rep(i,n) if(a[i]<0) ans.emplace_back(mai,i+1);
    rep(i,n-1) ans.emplace_back(i+1,i+2);
  }else{
    rep(i,n) if(a[i]>0) ans.emplace_back(mii,i+1);
    rep(i,n-1) ans.emplace_back(n-i,n-i-1);
  }
  int m = ans.size();
  cout << m << endl;
  rep(i,m) cout << ans[i].first << " " << ans[i].second << endl;
}