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;
#define rep(i,n) for(int i=0;i<n;i++)
#define rep2(i,a,b) for(int i=(a);i<(b);++i)

template<class T> inline void chmin(T& a, T b) {if (a > b) a = b;}
template<class T> inline void chmax(T& a, T b) {if (a < b) a = b;}

int main() {
  int N;
  cin >> N;
  vector<int> a(N);
  rep(i,N) cin >> a[i];
  cout << N+N-2 << endl;
  if (*max_element(a.begin(),a.end()) >= abs(*min_element(a.begin(),a.end()))) {
    auto ix = max_element(a.begin(),a.end());
    rep(i,N){if (ix-a.begin()+1!=i+1)
      cout << ix-a.begin()+1 << ' ' << i+1 << endl;
    }
    rep2(i,1,N) {
      cout << i << ' ' << i+1 << endl;
    }
  }
  else {
    auto ix = min_element(a.begin(),a.end());
    rep(i,N){if (ix-a.begin()+1!=i+1)
      cout << ix-a.begin()+1 << ' ' << i+1 << endl;
    }
    for (int i=N;i>1;i--) {
      cout << i << ' ' << i-1 << endl;
    }
  }

} 