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;
template <typename T> void read(T &t) {
	t=0; char ch=getchar(); int f=1;
	while ('0'>ch||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	do {(t*=10)+=ch-'0';ch=getchar();} while ('0'<=ch&&ch<='9'); t*=f;
}
typedef long long ll;
const ll INF=(9e18);
const int maxn=(2e5)+10;
int n,a[maxn];
ll dp[maxn][5],ans=INF;
int op(int i,int j) {
	if (j==1||j==3) {
		if (a[i]==0) return 2;
		return a[i]&1;
	}
	if (j==2) return a[i]&1^1;
	return a[i];
}
int main() {
	//freopen("1.txt","r",stdin);
	read(n);
	for (int i=1;i<=n;i++) read(a[i]);
	for (int i=1;i<=n;i++)
	for (int j=0;j<=4;j++) {
		dp[i][j]=INF;
		for (int k=0;k<=j;k++)
			dp[i][j]=min(dp[i][j],dp[i-1][k]);
		dp[i][j]+=op(i,j);
		if (i==n) ans=min(ans,dp[i][j]);
	}
	printf("%lld\n",ans);
	return 0;
}