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
#define chkmin(x, y) (x = min(x, y))
const int L = 200010;
long long dp[L][5], a[L];
int n;
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
//	function<void(int,int) > dfs = [&](int x, int y)
//	{
//		
//	};
	cin >> n;
	for (int i = 1; i <= n; ++i)
		cin >> a[i];
	memset(dp, 0x3f, sizeof(dp));
	dp[0][0] = dp[0][1] = dp[0][2] = dp[0][3] = dp[0][4] = 0;
	for (int i = 1; i <= n; ++i)
	{
        long long cur = dp[i - 1][0];
		dp[i][0] = dp[i - 1][0] + a[i];
		dp[i][1] = chkmin(cur, dp[i - 1][1]) + (a[i] == 0 ? 2 : (a[i] & 1));
		dp[i][2] = chkmin(cur, dp[i - 1][2]) + ((a[i] + 1) & 1);
		dp[i][3] = chkmin(cur, dp[i - 1][3]) + (a[i] == 0 ? 2 : (a[i] & 1));
		dp[i][4] = chkmin(cur, dp[i - 1][4]) + a[i];
	}
	long long ans = (LL)1e18;
	for (int i = 0; i <= 4; ++i)
		ans = min(ans, dp[n][i]);
	cout << ans;
    return 0;
}
