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
#!/usr/bin/env python3
import sys

sys.setrecursionlimit(10 ** 8)
ni = lambda: int(sys.stdin.readline())
nm = lambda: map(int, sys.stdin.readline().split())
nl = lambda: list(nm())
ns = lambda: sys.stdin.readline().rstrip()

L = ni()
# odd => 0, even => 1
A = [ni() for _ in range(L)]
assert len(A) == L

INF = 1 << 60
S = 5  # num of states


def solve():
    dp = [[INF] * S for _ in range(L + 1)]
    for s in range(S):
        dp[0][s] = 0
    for i in range(1, L + 1):
        a = A[i - 1]
        dp[i][0] = dp[i - 1][0] + a
        tm = min(dp[i - 1][0], dp[i - 1][1])
        dp[i][1] = tm + (2 if a == 0 else a % 2)
        tm = min(tm, dp[i - 1][2])
        dp[i][2] = tm + (1 - a % 2)
        tm = min(tm, dp[i - 1][3])
        dp[i][3] = tm + (2 if a == 0 else a % 2)
        tm = min(tm, dp[i - 1][4])
        dp[i][4] = tm + a
    # print(*dp, sep="\n", file=sys.stderr)
    return min(dp[L])


print(solve())
