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 ll long long

int main(void) {
    int n, x;
    while(~scanf("%d", &n)) {
        int abs_mx = 0, idx = -1;
        for(int i = 1; i <= n; ++i) {
            scanf("%d", &x);
            if(abs(abs_mx) < abs(x)) {
                abs_mx = x;
                idx = i;
            }
        }
        if(idx == -1) {
            puts("0");
            continue;
        }
        printf("%d\n", (n << 1) - 1);
        for(int i = 1; i <= n; ++i)
            printf("%d %d\n", idx, i);
        if(abs_mx > 0) {
            for(int i = 1; i < n; ++i)
                printf("%d %d\n", i, i + 1);
            } else {
                for(int i = n; i > 1; --i)
                    printf("%d %d\n", i, i - 1);
            }
    }
}
