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;
#define REP(i,a) for(int i = 0; i < (a); i++)
#define ALL(a) (a).begin(),(a).end()
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1e9;
const long long LINF = 1e18;
const long long MOD = 1e9 + 7;

signed main(){
    int n;
    cin >> n;
    vector<P> vec;
    int a[n];
    REP(i,n){
        cin >> a[i];
        vec.emplace_back(a[i], i);
    }
    sort(ALL(vec));
    cout << n * 2 - 1 << endl;
    if(vec[n - 1].first >= abs(vec[0].first)){
        for(int i = 1; i <= n; i++){
            cout << vec[n - 1].second + 1 << " " << i << endl;
        }
        for(int i = 1; i < n; i++){
            cout << i << " " << i + 1 << endl;
        }
    }else{
        for(int i = 1; i <= n; i++){
            cout << vec[0].second + 1 << " " << i << endl;
        }
        for(int i = n; i > 1; i--){
            cout << i << " " << i - 1 << endl;
        }
    }
    return 0;
}