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
// Example program

#include <bits/stdc++.h>
using namespace std;

typedef pair<int,int> ii;
typedef vector<ii> vii;

int main() {
    int n;
    cin >> n;
    vector<int> arr(n);
    for(int i = 0; i < n; i++) cin >> arr[i];
    int index = 0;
    int number = 0;
    for(int i = 0; i < n; i++) {
        if(abs(arr[i]) > number) {
            index = i;
            number = abs(arr[i]);
        }    
    }
    vii ans;
    
    for(int i = 0; i < n; i++) {
        ans.push_back(ii(index,i));
        arr[i] += arr[index];
    }
    if(arr[index] < 0) {
        for(int i = n-2; i >= 0; i--) {
            ans.push_back(ii(i+1,i));    
        }
    } else { 
        for(int i = 1; i < n; i++) {
            ans.push_back(ii(i-1,i));
        }    
    }
    cout << ans.size() << endl;
    for(int i = 0; i < ans.size(); i++ ){
        cout << ans[i].first+1 << " " << ans[i].second+1 << endl;
    }
}
