#include<iostream>
using namespace std;
typedef long long ll;

int main(){
    ll a, b, c, d, g;
    cin >> a >> b >> g >> c >> d >> g >> g;
    ll ans = (ll)b+a/2*2+c/2*2+d/2*2;
    switch(a%2+c%2+d%2){
        case 3:
            ans += 3;
            break;
        case 2:
            ans += (a%2==0&&a>0)||(c%2==0&&c>0)||(d%2==0&&d>0);
            break;
    }
    cout << ans << endl;
    return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

A tetromino is a figure formed by joining four squares edge to edge. We will refer to the following seven kinds of tetromino as I-, O-, T-, J-, L-, S- and Z-tetrominos, respectively:

a60bcb8e9e8f22e3af51049eda063392.png

Snuke has many tetrominos. The number of I-, O-, T-, J-, L-, S- and Z-tetrominos in his possession are a_I, a_O, a_T, a_J, a_L, a_S and a_Z, respectively. Snuke will join K of his tetrominos to form a rectangle that is two squares tall and 2K squares wide. Here, the following rules must be followed:

* When placing each tetromino, rotation is allowed, but reflection is not.
* Each square in the rectangle must be covered by exactly one tetromino.
* No part of each tetromino may be outside the rectangle.



Snuke wants to form as large a rectangle as possible. Find the maximum possible value of K.

Constraints

* 0≤a_I,a_O,a_T,a_J,a_L,a_S,a_Z≤10^9
* a_I+a_O+a_T+a_J+a_L+a_S+a_Z≥1

Input

The input is given from Standard Input in the following format:


a_I a_O a_T a_J a_L a_S a_Z


Output

Print the maximum possible value of K. If no rectangle can be formed, print `0`.

Examples

Input

2 1 1 0 0 0 0


Output

3


Input

0 0 10 0 0 0 0


Output

0

### Response

```cpp
#include<iostream>
using namespace std;
typedef long long ll;

int main(){
    ll a, b, c, d, g;
    cin >> a >> b >> g >> c >> d >> g >> g;
    ll ans = (ll)b+a/2*2+c/2*2+d/2*2;
    switch(a%2+c%2+d%2){
        case 3:
            ans += 3;
            break;
        case 2:
            ans += (a%2==0&&a>0)||(c%2==0&&c>0)||(d%2==0&&d>0);
            break;
    }
    cout << ans << endl;
    return 0;
}

```