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 l;
long long a[200005];
long long dp[200005][5];
constexpr long long big = 1000000000000000;
int main()
{
	std::cin >> l;
	for (int i = 0; i < l; ++i)
		std::cin >> a[i];

	for (int i = 0; i <= l + 1; ++i)
		for (int j = 0; j <= 4; ++j)
			dp[i][j] = big;

	dp[0][0] = 0;
	for (int i = 0; i < l; ++i) {
		dp[i + 1][0] = dp[i][0] + a[i];
		dp[i + 1][1] = std::min(dp[i][0], dp[i][1]) + (a[i] ? a[i] % 2 : 2);
		dp[i + 1][2] = std::min({ dp[i][0],dp[i][1],dp[i][2] }) + (a[i] ? !(a[i] % 2) : 1);
		dp[i + 1][3] = std::min({ dp[i][0],dp[i][1],dp[i][2],dp[i][3] }) + (a[i] ? a[i] % 2 : 2);
		dp[i + 1][4] = std::min({ dp[i][0],dp[i][1],dp[i][2],dp[i][3],dp[i][4] }) + a[i];
	}

	long long ans = big;
	for (int i = 0; i <= 4; ++i)
		ans = std::min(ans, dp[l][i]);

	std::cout << ans << std::endl;
	return 0;
}
