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 itertools

l = int(input())
a = [int(input()) for _ in range(l)]

e_cs = [int(not (v % 2)) for v in a]
o_cs = [2 if v == 0 else v % 2 for v in a]

cs = [0] + list(itertools.accumulate(a))
cost = [a, o_cs, e_cs, o_cs, a]

# dp = [[None] * 5 for _ in range(l + 1)]
# def dfs(i=0, t=0):
#     if dp[i][t] is not None:
#         return dp[i][t]
#
#     if i == l:
#         dp[i][t] = 0
#         return 0
#     if t == 4:
#         dp[i][t] = cs[l] - cs[i]
#         return dp[i][t]
#     dp[i][t] = min(dfs(i + 1, t) + cost[t][i], dfs(i, t + 1))
#     return dp[i][t]

dp = [[None] * 5 for _ in range(l + 1)]
dp[l] = [0] * 5

for i in reversed(range(l)):
    for t in reversed(range(5)):
        if t == 4:
            dp[i][t] = cs[l] - cs[i]
            continue
        dp[i][t] = min(dp[i + 1][t] + cost[t][i], dp[i][t + 1])

print(dp[0][0])
