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

"""
Writer: SPD_9X2
https://atcoder.jp/contests/yahoo-procon2019-qual/tasks/yahoo_procon2019_qual_d

startとGoalを決めると、その間は奇数回、それ以外は偶数回通ることとなる
また、始点と終点は交換可能なので、始点が左にある方としてよい

始点・終点間は最小１で、好きな位置を2ずつ増やすことが可能
始点から左、終点より右は0があればそれ以降先は全て0にせざるを得ない
それ以外は2,4,6…と2ずつ好きに増やすことが可能

→大体情報が出そろってきた

もし終点と始点が全探索可能だとしたら？
始点より左の最適化 = 終点より右はどうするんだ？
→切れ目を全探索すれば行ける　始点・終点間は自明

高速化→dpだろうなぁ
フェーズごとに分ける

左のずっと0フェーズ→偶数フェーズ→(始点)→奇数フェーズ→(終点)→偶数フェーズ→0フェーズ
の5段階ごとに最小値を持っておいてあげればよい
あとは適当にdp(フェーズは戻っちゃだめだけど飛ばしても良いことに注意)
→解けた！！！
"""

dp = [0] * 5

L = int(input())

for i in range(L):

    ndp = [float("inf")] * 5

    A = int(input())

    ndp[0] = dp[0] + A

    if A == 0:
        D = 2
    else:
        D = A % 2
    ndp[1] = min(dp[0],dp[1]) + D
    ndp[2] = min(dp[0],dp[1],dp[2]) + (A+1)%2
    ndp[3] = min(dp[0],dp[1],dp[2],dp[3]) + D
    ndp[4] = min(dp) + A

    dp = ndp

print (min(dp))