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 all(V) V.begin(),V.end()
#define ll long long
#define MOD 1000000007

ll dp[5][200010];//02120

int main(){
	int L;
	cin >> L;
	vector<ll> A(L), B(L);
	for (int i = 0;i < L;i++) {
		cin >> A[i];
		if (A[i] == 0)B[i] = 0;
		else if (A[i] & 1)B[i] = 1;
		else B[i] = 2;
	}

	dp[0][0] = dp[4][0] = A[0];
	dp[1][0] = dp[3][0] = abs(2 - B[0]);
	dp[2][0] = abs(1 - B[0]);

	for (int i = 1;i < L;i++) {
		dp[0][i] = dp[0][i - 1] + A[i];
		dp[1][i] = min(dp[1][i - 1], dp[0][i - 1]) + abs(2 - B[i]);
		dp[2][i] = min({ dp[2][i - 1],dp[1][i - 1],dp[0][i - 1] }) + abs(1 - B[i]);
		dp[3][i] = min({ dp[3][i - 1],dp[2][i - 1],dp[1][i - 1],dp[0][i - 1] }) + abs(2 - B[i]);
		dp[4][i] = min({ dp[4][i - 1],dp[3][i - 1],dp[2][i - 1],dp[1][i - 1],dp[0][i - 1] }) + A[i];
	}

	cout << min({ dp[4][L - 1],dp[3][L - 1],dp[2][L - 1],dp[1][L - 1],dp[0][L - 1] }) << endl;
}