#include<bits/stdc++.h>
using namespace std;
int main() {
    int N;
    cin >> N;
    vector<int>A(N);
    int sum = 0,j = 0;
    for(int i = 0; i < N; i++) {
        cin >> A[i];
        if(abs(A[i]) > sum) {
            j = i;
            sum = abs(A[i]);
        }
    }
    cout << 2*N << endl;
    if(A[j] < 0) {
        cout << j+1 << " " << N << endl;
        cout << j+1 << " " << N << endl;
        for(int i = N; i >= 2; i--) {
            cout << i << " " << i-1 << endl;
            cout << i << " " << i-1 << endl;
        }
    }
    else {
        cout << j+1 << " " << 1 << endl;
        cout << j+1 << " " << 1 << endl;
        for(int i = 1; i <= N-1; i++) {
            cout << i << " " << i+1 << endl;
            cout << i << " " << i+1 << endl;
        }
    }
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
int main() {
    int N;
    cin >> N;
    vector<int>A(N);
    int sum = 0,j = 0;
    for(int i = 0; i < N; i++) {
        cin >> A[i];
        if(abs(A[i]) > sum) {
            j = i;
            sum = abs(A[i]);
        }
    }
    cout << 2*N << endl;
    if(A[j] < 0) {
        cout << j+1 << " " << N << endl;
        cout << j+1 << " " << N << endl;
        for(int i = N; i >= 2; i--) {
            cout << i << " " << i-1 << endl;
            cout << i << " " << i-1 << endl;
        }
    }
    else {
        cout << j+1 << " " << 1 << endl;
        cout << j+1 << " " << 1 << endl;
        for(int i = 1; i <= N-1; i++) {
            cout << i << " " << i+1 << endl;
            cout << i << " " << i+1 << endl;
        }
    }
}
```