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>

int ri() {
	int n;
	scanf("%d", &n);
	return n;
}

int calc2(int n) {
	if (!n) return 2;
	else return n & 1;
}

int calc1(int n) {
	return !(n & 1);
}

int main() {
	int n = ri();
	int a[n];
	for (int i = 0; i < n; i++) a[i] = ri();
	
	int64_t dp[n + 1][5];
	for (int i = 0; i <= n; i++)
		for (int j = 0; j < 5; j++) dp[i][j] = 1000000000000000000;
	dp[0][0] = 0;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < 5; j++) {
			int64_t next;
			if (j == 0 || j == 4) next = dp[i][j] + a[i];
			else if (j == 1 || j == 3) next = dp[i][j] + calc2(a[i]);
			else next = dp[i][j] + calc1(a[i]);
			dp[i + 1][j] = std::min(dp[i + 1][j], next);
			if (j < 4) dp[i][j + 1] = std::min(dp[i][j + 1], dp[i][j]);
		}
	}
	std::cout << *std::min_element(dp[n], dp[n] + 5) << std::endl;
	
	return 0;
}
