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;

template <class T>
inline bool uin(T& a, T b) { return a > b ? (a = b, true) : false; }
template <class T>
inline bool uax(T& a, T b) { return a < b ? (a = b, true) : false; }

const int nax = 2e5 + 10;

long long dp[nax][5];
long long a[nax];
int N;

long long get(long long x, int c) {
	if (c == 0 || c == 4) {
		return x;
	}
	if (c == 1 || c == 3) {
		if (x == 0) return 2;
		return (x % 2);
	}
	return (x + 1LL) % 2;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	memset(dp, 57, sizeof(dp));
	for (int i = 0; i < 5; ++i) dp[0][i] = 0;

	cin >> N;
	for (int i = 0; i < N; ++i) cin >> a[i];

	for (int i = 0; i < N; ++i)
		for (int j = 0; j < 5; ++j)
			for (int k = j; k < 5; ++k)
				uin(dp[i + 1][k], dp[i][j] + get(a[i], k));

	long long ans = 1e18L;
	for (int i = 0; i < 5; ++i) uin(ans, dp[N][i]);

	cout << ans << '\n';
}