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 ALL(c) (c).begin(),(c).end()
#define EACH(i,c) for(auto i=(c).begin();i!=(c).end();++i)
using namespace std;
#define int long long

const int INF = (int)1e18;

signed main(){
	int L;
	cin >> L;
	vector<int> A(L + 2);
	for(int i = 1; i <= L; ++i) {
		cin >> A[i];
	}
	vector<vector<int>> dp(L + 1, vector<int>(5, INF));
	dp[0].assign(5, 0);
	for(int i = 1; i <= L; ++i) {
		int r = dp[i - 1][0];

		dp[i][0] = r + A[i];

		r = min(r, dp[i - 1][1]);
		
		int a = 0;
		if (A[i] == 0) {
			a = 2;
		} else if(A[i] % 2 == 1) {
			a = 1;
		}
		dp[i][1] = r + a;

		r = min(r, dp[i - 1][2]);
		int b = A[i] % 2 == 0 ? 1 : 0;
		dp[i][2] = r + b;

		r = min(r, dp[i - 1][3]);
		dp[i][3] = r + a;

		r = min(r, dp[i - 1][4]);
		dp[i][4] = r + A[i];
	}

	int res = dp[L][0];
	for(auto a: dp[L])
		res = min(res, a);
	cout << res << endl;
	return 0;
}
