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
import sys
input = sys.stdin.readline

L = int(input())
A = [int(input()) for _ in range(L)]
dpl = [[10**18]*2 for _ in range(L+1)]
dpl[0][1] = 0
acc = 0

for i in range(1, L+1):
    dpl[i][0] = min(dpl[i-1][0], dpl[i-1][1])+(1 if A[i-1]%2==0 else 0)
    acc += A[i-1]
    dpl[i][1] = acc
    
    if A[i-1]%2==0:
        if A[i-1]>=2:
            dpl[i][1] = min(dpl[i][1], dpl[i-1][1])
        else:
            dpl[i][1] = min(dpl[i][1], dpl[i-1][1]+2)
    else:
        dpl[i][1] = min(dpl[i][1], dpl[i-1][1]+1)
    
acc = 0
dpr = [[10**18]*2 for _ in range(L+1)]
dpr[L][1] = 0

for i in range(L-1, -1, -1):
    dpr[i][0] = min(dpr[i+1][0], dpr[i+1][1])+(1 if A[i]%2==0 else 0)
    acc += A[i]
    dpr[i][1] = acc
    
    if A[i]%2==0:
        if A[i]>=2:
            dpr[i][1] = min(dpr[i][1], dpr[i+1][1])
        else:
            dpr[i][1] = min(dpr[i][1], dpr[i+1][1]+2)
    else:
        dpr[i][1] = min(dpr[i][1], dpr[i+1][1]+1)

ans = 10**18

for i in range(L+1):
    ans = min(ans, dpl[i][0]+dpr[i][1], dpl[i][1]+dpr[i][0], dpl[i][1]+dpr[i][1])

print(ans)