#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

vector<ll> f(const vector<ll> &a) {
	const int n = (int) a.size();
	vector<ll> dp(n + 1, 0);
	ll sum = 0;
	for (int i = 1; i <= n; ++i) {
		const ll x = a[i - 1];
		sum += x;
		dp[i] = min(sum, dp[i - 1] + (x - 2 < 0 ? 2 - x : x & 1));
	}
	return dp;
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);

	int n;
	cin >> n;
	vector<ll> a(n);
	for (ll &x : a)
		cin >> x;

	vector<ll> dp = f(a);

	for (int i = 1; i <= n; ++i) {
		const ll x = a[i - 1];
		dp[i] = min(dp[i], dp[i - 1] + (x == 0 ? 1 : (x - 1) & 1));
	}

	reverse(a.begin(), a.end());
	vector<ll> dp2 = f(a);

	ll ans = 1LL << 60;
	for (int i = 0; i <= n; ++i) {
		ans = min(ans, dp[i] + dp2[n - i]);
	}
	cout << ans;
	return 0;
}

### Prompt

Generate a Cpp solution to 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
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

vector<ll> f(const vector<ll> &a) {
	const int n = (int) a.size();
	vector<ll> dp(n + 1, 0);
	ll sum = 0;
	for (int i = 1; i <= n; ++i) {
		const ll x = a[i - 1];
		sum += x;
		dp[i] = min(sum, dp[i - 1] + (x - 2 < 0 ? 2 - x : x & 1));
	}
	return dp;
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);

	int n;
	cin >> n;
	vector<ll> a(n);
	for (ll &x : a)
		cin >> x;

	vector<ll> dp = f(a);

	for (int i = 1; i <= n; ++i) {
		const ll x = a[i - 1];
		dp[i] = min(dp[i], dp[i - 1] + (x == 0 ? 1 : (x - 1) & 1));
	}

	reverse(a.begin(), a.end());
	vector<ll> dp2 = f(a);

	ll ans = 1LL << 60;
	for (int i = 0; i <= n; ++i) {
		ans = min(ans, dp[i] + dp2[n - i]);
	}
	cout << ans;
	return 0;
}

```