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)]
INF = 10**18
dp = [[INF] * 5 for _ in range(L+1)]
dp[0][0] = 0
# none, even, odd, even, none
for i in range(L):
    for j in range(5):
        for k in range(j, 5):
            if k in (0, 4): # none
                dp[i+1][k] = min(dp[i+1][k], dp[i][j] + A[i])
            if k in (1, 3): # even
                if A[i] == 0:
                    dp[i+1][k] = min(dp[i+1][k], dp[i][j] + 2)
                else:
                    dp[i+1][k] = min(dp[i+1][k], dp[i][j] + (A[i] % 2))
            if k == 2: # odd
                if A[i] == 0:
                    dp[i+1][k] = min(dp[i+1][k], dp[i][j] + 1)
                else:
                    dp[i+1][k] = min(dp[i+1][k], dp[i][j] + 1 - (A[i] % 2))
print(min(dp[L]))