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
#include <bits/stdc++.h>
using namespace std;
using lld = int64_t;
const int N = 200000 + 5;

int a[ N ];
lld dp[ N ][ 5 ];

int main() {
	int n; scanf( "%d", &n );
	for ( int i = 1 ; i <= n ; ++ i )
		scanf( "%d", a + i );
	for ( int i = 1 ; i <= n ; ++ i ) {
		lld mi = dp[ i - 1 ][ 0 ];
		dp[ i ][ 0 ] = mi + a[ i ];
		mi = min( mi, dp[ i - 1 ][ 1 ] );
		dp[ i ][ 1 ] = mi + ( a[ i ] ? ( a[ i ] & 1 ) : 2 );
		mi = min( mi, dp[ i - 1 ][ 2 ] );
		dp[ i ][ 2 ] = mi + ( ( a[ i ] + 1 ) & 1 );
		mi = min( mi, dp[ i - 1 ][ 3 ] );
		dp[ i ][ 3 ] = mi + ( a[ i ] ? ( a[ i ] & 1 ) : 2 );
		mi = min( mi, dp[ i - 1 ][ 4 ] );
		dp[ i ][ 4 ] = mi + a[ i ];
	}
	lld mi = dp[ n ][ 0 ];
	for ( int i = 1 ; i < 5 ; ++ i )
		mi = min( mi, dp[ n ][ i ] );
	printf( "%" PRId64 "\n", mi );
	return 0;
}