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
L = int(input())
B = [int(input()) for _ in range(L)]

A = []
for b in B:
    if b % 2:
        A.append(1)
    else:
        A.append(0 if b == 0 else 2)

dpLBack = [10**10] * (L + 1)
dpLBack[0] = 0
cnt = 0
for i, a in enumerate(A, start=1):
    cnt += B[i - 1]
    dpLBack[i] = min(dpLBack[i - 1] + 2 - a, cnt)

dpL = [10**10] * (L + 1)
dpL[0] = 0
cnt = 0
for i, a in enumerate(A, start=1):
    cnt += B[i - 1]
    dpL[i] = min(dpL[i - 1] + (a != 1), dpLBack[i - 1] + (a == 0), cnt)

dpRBack = [10**10] * (L + 1)
dpRBack[0] = 0
cnt = 0
for i, a in enumerate(A[:: -1], start=1):
    cnt += B[-i]
    dpRBack[i] = min(dpRBack[i - 1] + 2 - a, cnt)

dpR = [10**10] * (L + 1)
dpR[0] = 0
cnt = 0
for i, a in enumerate(A[:: -1], start=1):
    cnt += B[-i]
    dpR[i] = min(dpR[i - 1] + (a != 1), dpRBack[i - 1] + (a == 0), cnt)

dpR = dpR[:: -1]
dpRBack = dpRBack[::-1]

ans = min(dpL[L], dpR[0])
for mid in range(L + 1):
    ans = min(ans, dpL[mid] + dpRBack[mid], dpLBack[mid] + dpR[mid], dpLBack[mid] + dpRBack[mid])

print(ans)
