Snuke stands on a number line. He has L ears, and he will walk along the line continuously under the following conditions:

* He never visits a point with coordinate less than 0, or a point with coordinate greater than L.
* He starts walking at a point with integer coordinate, and also finishes walking at a point with integer coordinate.
* He only changes direction at a point with integer coordinate.



Each time when Snuke passes a point with coordinate i-0.5, where i is an integer, he put a stone in his i-th ear.

After Snuke finishes walking, Ringo will repeat the following operations in some order so that, for each i, Snuke's i-th ear contains A_i stones:

* Put a stone in one of Snuke's ears.
* Remove a stone from one of Snuke's ears.



Find the minimum number of operations required when Ringo can freely decide how Snuke walks.

Constraints

* 1 \leq L \leq 2\times 10^5
* 0 \leq A_i \leq 10^9(1\leq i\leq L)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


L
A_1
:
A_L


Output

Print the minimum number of operations required when Ringo can freely decide how Snuke walks.

Examples

Input

4
1
0
2
3


Output

1


Input

8
2
0
0
2
1
3
4
1


Output

3


Input

7
314159265
358979323
846264338
327950288
419716939
937510582
0


Output

1
from sys import stdin
from itertools import repeat
def main():
    n = int(stdin.readline())
    a = [int(stdin.readline()) for _ in xrange(n)]
    while a and not a[-1]:
        a.pop()
    a.reverse()
    while a and not a[-1]:
        a.pop()
    n = len(a)
    p = [0] * n
    q = [0] * n
    for i in xrange(n):
        if not a[i]:
            p[i] = 1
            q[i] = 2
        else:
            p[i] = 1 - a[i] % 2
            q[i] = a[i] % 2
    l = [(0, 0, 0)] * (n + 1)
    r = [(0, 0, 0)] * (n + 1)
    for i in xrange(n):
        l[i+1] = (l[i][0] + a[i], min(l[i][0], l[i][1]) + q[i], min(l[i]) + p[i])
    for i in xrange(n - 1, -1, -1):
        r[i] = (r[i+1][0] + a[i], min(r[i+1][0], r[i+1][1]) + q[i], min(r[i+1]) + p[i])
    ans = min(min(l[i]) + min(r[i]) for i in xrange(n + 1))
    print ans
main()
