//#pragma GCC optimize(2)
// Happy TLE and WA every day!
// by: zxb the vegetable chicken
#include<bits/stdc++.h>
#define rep(i,n)        for(int i = 0; i < n; i++)
#define INF             0x3f3f3f3f
using namespace std;

// 0 2 1 2 0
int a[200200];
int dp[200200][5];
int n;

signed main(){
	ios::sync_with_stdio(false);
	cin >> n;
	rep(i, n) cin >> a[i];
	for(int i = 1; i <= n; i++){
		int mn = INF;
		int x = a[i - 1];
		mn = min(mn, dp[i - 1][0]);
		dp[i][0] = mn + x;
		mn = min(mn, dp[i - 1][1]);
		dp[i][1] = mn + ( (!x) ? 2 : (x&1) );
		mn = min(mn, dp[i - 1][2]);
		dp[i][2] = mn + !(x & 1);
		mn = min(mn, dp[i - 1][3]);
		dp[i][3] = mn + ( (!x) ? 2 : (x&1) );
		mn = min(mn, dp[i - 1][4]);
		dp[i][4] = mn + x;
	}
	int ans = INF;
	rep(i, 5) ans = min(ans, dp[n][i]);
	cout << ans << endl;
	return 0;
}
### Prompt

Create a solution in CPP for the following problem:

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

### Response

```cpp
//#pragma GCC optimize(2)
// Happy TLE and WA every day!
// by: zxb the vegetable chicken
#include<bits/stdc++.h>
#define rep(i,n)        for(int i = 0; i < n; i++)
#define INF             0x3f3f3f3f
using namespace std;

// 0 2 1 2 0
int a[200200];
int dp[200200][5];
int n;

signed main(){
	ios::sync_with_stdio(false);
	cin >> n;
	rep(i, n) cin >> a[i];
	for(int i = 1; i <= n; i++){
		int mn = INF;
		int x = a[i - 1];
		mn = min(mn, dp[i - 1][0]);
		dp[i][0] = mn + x;
		mn = min(mn, dp[i - 1][1]);
		dp[i][1] = mn + ( (!x) ? 2 : (x&1) );
		mn = min(mn, dp[i - 1][2]);
		dp[i][2] = mn + !(x & 1);
		mn = min(mn, dp[i - 1][3]);
		dp[i][3] = mn + ( (!x) ? 2 : (x&1) );
		mn = min(mn, dp[i - 1][4]);
		dp[i][4] = mn + x;
	}
	int ans = INF;
	rep(i, 5) ans = min(ans, dp[n][i]);
	cout << ans << endl;
	return 0;
}
```