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;

int main() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (auto& e : a) cin >> e;
    auto minitr = min_element(begin(a), end(a)), maxitr = max_element(begin(a), end(a));
    int imin = *minitr, imax = *maxitr;
    vector<pair<int, int>> ans;

    if (imin < 0 && imax > 0) {
        if (imin + imax > 0) {
            for (int i = 1; i <= n; i++) ans.emplace_back(maxitr - begin(a) + 1, i);
        } else {
            for (int i = 1; i <= n; i++) ans.emplace_back(minitr - begin(a) + 1, i);
        }
    }

    if (imin + imax > 0) {
        for (int i = 1; i + 1 <= n; i++) ans.emplace_back(i, i + 1);
    } else {
        for (int i = n; i - 1 >= 1; i--) ans.emplace_back(i, i - 1);
    }

    cout << ans.size() << endl;
    for (auto& e : ans) cout << e.first << " " << e.second << endl;
}
