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>
#define REP(i, n) for (int i = 0; (i) < int(n); ++ (i))
#define REP_R(i, n) for (int i = int(n) - 1; (i) >= 0; -- (i))
#define ALL(x) begin(x), end(x)
using namespace std;

int main() {
    // input
    int n; scanf("%d", &n);
    vector<int> a(n); REP (i, n) scanf("%d", &a[i]);
    // solve
    int argmin_a = min_element(ALL(a)) - a.begin();
    int argmax_a = max_element(ALL(a)) - a.begin();
    vector<pair<int, int> > result;
    if (abs(a[argmin_a]) <= abs(a[argmax_a])) {
        REP (i, n) if (i != argmax_a) result.emplace_back(argmax_a, i);
        REP (i, n - 1) result.emplace_back(i, i + 1);
    } else {
        REP (i, n) if (i != argmin_a) result.emplace_back(argmin_a, i);
        REP_R (i, n - 1) result.emplace_back(i + 1, i);
    }
    // output
    printf("%d\n", int(result.size()));
    for (auto op : result) {
        int x, y; tie(x, y) = op;
        printf("%d %d\n", x + 1, y + 1);
    }
    return 0;
}
