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;
using ll = long long;
#define rep(i, n) for (int i = 0; i < (n); i++)

int main() {
    int n;
    cin >> n;
    vector<int> a(n);
    int amax = 0, amin = 0;
    rep(i, n){
        cin >> a[i];
        if(a[i] < a[amin]){
            amin = i;
        }
        if(a[i] > a[amax]){
            amax = i;
        }
    }
    vector<vector<int>> operations;
    if (abs(a[amax]) >= abs(a[amin])) {
        rep(i,n){
            if(i == amax) continue;
            operations.push_back({amax, i});
        }
        for(int i = 0; i < n-1; i++) operations.push_back({i, i + 1});
    } else {
        rep(i,n){
            if(i == amin) continue;
            operations.push_back({amin, i});
        }
        for(int i = n-1; i > 0; i--) operations.push_back({i, i-1});
    }

    cout << operations.size() << endl;
    for(auto v : operations){
        cout << v[0]+1 << " " << v[1]+1 << endl;
    }
}