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 ll long long
using namespace std;
ll MOD = 1e9 + 7;
ll A,B,H, N, M, L, K, W, X, Y, Z;
ll ans = 0;
int main() {
	cin >> L;
	vector<ll>A(L + 1);
	vector<vector<ll>>dp(5, vector<ll>(L + 1));
	for (int t = 0; t <= 4; t++)dp[t][0] = 0;
	for (int i = 1; i <= L; i++) {
		cin >> A[i];
		ll min1 = min(dp[0][i - 1], dp[1][i - 1]);
		ll min2 = min(dp[2][i - 1], min1);
		ll min3 = min(dp[3][i - 1], min2);
		ll min4 = min(dp[4][i - 1], min3);
		dp[0][i] = dp[0][i - 1] + A[i];
		dp[4][i] = min4 + A[i];
		if (A[i] & 1) {
			dp[1][i] = min1+ 1;
			dp[2][i] = min2;
			dp[3][i] = min3+ 1;
		}
		else if (!(A[i] & 1) && A[i] > 0) {
			dp[1][i] = min1;
			dp[2][i] = min2 + 1;
			dp[3][i] = min3;
		}
		else {
			dp[1][i] = min1 + 2;
			dp[2][i] = min2 + 1;
			dp[3][i] = min3 + 2;
		}
	}
	ans = dp[0][L];
	for (int t = 1; t <= 4; t++)ans = min(ans, dp[t][L]);
	cout << ans << endl;
}
