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
n = int(input())
a = [int(input()) for i in range(n)]
b = [0]*n
for i in range(n):
  if a[i]%2:
    b[i] = 1
  elif a[i]:
    b[i] = 2
a.append(0)
b.append(0)
n += 1
DP = [[[10**18 for i in range(3)] for j in range(3)] for k in range(n+1)]
for i in range(3):
  for j in range(3):
    DP[0][i][j] = 0
for i in range(1,n+1):
  for k in (0,2):
    DP[i][0][k] = min(DP[i][0][k],DP[i-1][0][k]+abs(a[i-1]))
  for l in (1,2):
    DP[i][0][2] = min(DP[i][0][2],DP[i-1][l][2]+abs(a[i-1]))
  DP[i][0][2] = min(DP[i][0][2],DP[i-1][2][1]+abs(a[i-1]))

  DP[i][1][2] = min(DP[i][1][2],DP[i-1][0][0]+abs(b[i-1]-1))
  DP[i][1][2] = min(DP[i][1][2],DP[i-1][1][2]+abs(b[i-1]-1))
  DP[i][1][2] = min(DP[i][1][2],DP[i-1][2][1]+abs(b[i-1]-1))

  for k in (1,2):
    DP[i][2][k] = min(DP[i][2][k],DP[i-1][2][k]+abs(b[i-1]-2))
  DP[i][2][2] = min(DP[i][2][2],DP[i-1][1][2]+abs(b[i-1]-2))
  DP[i][2][1] = min(DP[i][2][1],DP[i-1][0][0]+abs(b[i-1]-2))
print(min(*DP[n][0]))