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
def cost(place,zone):
    if zone == 0 or zone == 4:
        return A[place]
    elif zone == 1 or zone == 3:
        if A[place] == 0:
            return 2
        elif A[place] % 2 == 0:
            return 0
        else:
            return 1
    else:
        if A[place] % 2 == 0:
            return 1
        else:
            return 0

if __name__=='__main__':
    L = int(input())
    A = [int(input()) for i in range(L)]

    # dp配列の初期化
    dp = [[0 for _ in range(5)] for _ in range(L+10)]
    # 初項の設定
    for j in range(5):
        dp[0][j] = cost(0,j)
    # dp配列の更新
    for i in range(1,L):
        dp[i][0] = dp[i-1][0]+cost(i,0)
        for j in range(1,5):
            # 0〜jまでの最小に，i番目をjにするためのコストを足したものが最小
            dp[i][j] = min(min(dp[i-1][0:j+1])+cost(i,j),dp[i][j-1])
    # print(dp)
    print(min(dp[L-1]))