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;
#define ll long long

const int N = 2e5;
int n;
ll A[N + 87];
ll dp[N + 87][5];///all->#odd->#even->#odd->all

int iseven(ll k){
	return (k % 2 == 0) ? 1 : 0;
}
int isodd(ll k){
	if(k == 0)
		return 2;
	return 1 - iseven(k);
}

int main(){
	cin >> n;
	for(int i = 0; i < n; i++)
		cin >> A[i];
	ll now = A[0];
	dp[0][0] = now;
	dp[0][1] = isodd(now);
	dp[0][2] = iseven(now);
	dp[0][3] = isodd(now);
	dp[0][4] = now;

	for(int i = 1; i <= n; i++){
		now = A[i];
		ll tmp = dp[i - 1][0];
		dp[i][0] = tmp + now;
		tmp = min(tmp, dp[i - 1][1]);
		dp[i][1] = tmp + isodd(now);
		tmp = min(tmp, dp[i - 1][2]);
		dp[i][2] = tmp + iseven(now);
		tmp = min(tmp, dp[i - 1][3]);
		dp[i][3] = tmp + isodd(now);
		tmp = min(tmp, dp[i - 1][4]);
		dp[i][4] = tmp + now;
	}

	ll ans = dp[n][1];
	for(int i = 1; i < 5; i++)
		ans = min(ans, dp[n][i]);
	cout << ans << "\n";
}