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>
#define rep(i,k,n) for(int i=k;i<=n;i++)
#define rep2(i,k,n) for(int i=k;i>=n;i--)
using namespace std;
const int N=2e5+7;
typedef long long ll;
ll dp[2][2][N];//dp[l/r][][n]
int L,a[N];
int main(){
	memset(dp,0x3f,sizeof(dp));
	scanf("%d",&L);
	ll sum=0;
	dp[0][0][0]=dp[0][1][0]=dp[1][0][L+1]=dp[1][1][L+1]=0;
	rep(i,1,L){
		scanf("%d",&a[i]);
		sum+=a[i];
		dp[0][0][i]=min(sum, dp[0][0][i-1] + (a[i] ? (a[i]&1) : 2));
		dp[0][1][i]=min(dp[0][0][i], dp[0][1][i-1] + (a[i] ? !(a[i]&1) : 1));
	}
	sum=0;
	rep2(i,L,1){
		sum+=a[i];
		dp[1][0][i]=min(sum, dp[1][0][i+1] + (a[i] ? (a[i]&1) : 2));
		dp[1][1][i]=min(dp[1][0][i], dp[1][1][i+1] + (a[i] ? !(a[i]&1) : 1));
	}
	ll ans=0x3f3f3f3f3f3f3f3full;
	rep(i,1,L)ans=min(ans,dp[0][1][i-1] + dp[1][1][i]);
	printf("%lld\n",ans);
}