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
def a(xs):
    k = 0
    m = 0
    ops = []
    for i, x in enumerate(xs):
        if x > m:
            k = i
            m = x
    for i, x in enumerate(xs):
        if xs[i] < 0:
            xs[i] += m
            ops.append((k, i))
    for i in range(len(xs) - 1):
        if xs[i] > xs[i + 1]:
            xs[i + 1] += xs[i]
            ops.append((i, i + 1))
    return ops

def b(xs):
    k = 0
    m = 0
    ops = []
    for i, x in enumerate(xs):
        if x < m:
            k = i
            m = x
    for i, x in enumerate(xs):
        if xs[i] > 0:
            xs[i] += m
            ops.append((k, i))
    for i in range(len(xs) - 2, -1, -1):
        if xs[i] > xs[i + 1]:
            xs[i] += xs[i + 1]
            ops.append((i + 1, i))
    return ops
            
n = input()
xs = map(int, raw_input().split())
m = min(xs)
M = max(xs)
if abs(M) >= abs(m):
    ops = a(xs)
else:
    ops = b(xs)
print len(ops)
for a, b in ops:
    print a + 1, b + 1
