#include <iostream>
using namespace std;
typedef long long ll;
ll n, a[55], sei, fu, ma, mi;
int main(void){
    // Your code here!
    cin >> n;
    cout << 2 * n - 2 << endl;
    for (ll i = 1; i <= n; i++){
        cin >> a[i];
        if (a[i] > 0){
            if (a[i] >= sei){
                ma = i, sei = a[i];
            }
        } else {
            if (a[i] <= fu){
                mi = i, fu = a[i];
            }
        }
    }
    if (sei > -fu){
        for (ll i = 1; i <= n; i++){
            if (i == ma) continue;
            cout << ma << ' ' << i << endl;
        }
        for (ll i = 1; i < n; i++){
            cout << i << ' ' << i + 1 << endl;
        }
    } else {
        for (ll i = 1; i <= n; i++){
            if (i == mi) continue;
            cout << mi << ' ' << i << endl;
        }
        for (ll i = n; i > 1; i--){
            cout << i << ' ' << i - 1 << endl;
        }
    }
}

### Prompt

Please create a solution in CPP to the following problem:

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 <iostream>
using namespace std;
typedef long long ll;
ll n, a[55], sei, fu, ma, mi;
int main(void){
    // Your code here!
    cin >> n;
    cout << 2 * n - 2 << endl;
    for (ll i = 1; i <= n; i++){
        cin >> a[i];
        if (a[i] > 0){
            if (a[i] >= sei){
                ma = i, sei = a[i];
            }
        } else {
            if (a[i] <= fu){
                mi = i, fu = a[i];
            }
        }
    }
    if (sei > -fu){
        for (ll i = 1; i <= n; i++){
            if (i == ma) continue;
            cout << ma << ' ' << i << endl;
        }
        for (ll i = 1; i < n; i++){
            cout << i << ' ' << i + 1 << endl;
        }
    } else {
        for (ll i = 1; i <= n; i++){
            if (i == mi) continue;
            cout << mi << ' ' << i << endl;
        }
        for (ll i = n; i > 1; i--){
            cout << i << ' ' << i - 1 << endl;
        }
    }
}

```