Problem: 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

Solution (in PYTHON3):

Solution:
n = int(input())
a = list(map(int, input().split()))
#符号が同じ場合、累積和的な感じでいける。
#全部負か全部正しかない。
#全部正の場合は、一番おっきいやつを一番左に足し、左から累積和
#全部負の場合は、一番ちっさいやつを右に足し、右から累積和
#また、負と正が両方含まれている場合は、どっちかに統一する
ans = []
if max(a) * min(a) < 0:
    if abs(max(a)) > abs(min(a)):
        for i in range(n):
            if i != a.index(max(a)):
                ans.append([a.index(max(a))+1, i+1])
                a[i] += max(a)
    else:
        for i in range(n):
            if i != a.index(min(a)):
                ans.append([a.index(min(a))+1, i+1])
                a[i] += min(a)
if max(a) > 0:
    #全部正
    for i in range(1,n):
        ans.append([i,i+1])
else:
    #全部負
    for i in range(n-1,0,-1):
        ans.append([i+1,i])
print(len(ans))
for i in range(len(ans)):
    print("{} {}".format(*ans[i]))