Problem:
Score : 600 points

Problem Statement

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:

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.

Sample Input 1

2 1 1 0 0 0 0

Sample Output 1

3

One possible way to form the largest rectangle is shown in the following figure:

Sample Input 2

0 0 10 0 0 0 0

Sample Output 2

0

No rectangle can be formed.

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

use std::cmp::*;
use std::io::*;

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let I: usize = itr.next().unwrap().parse().unwrap();
    let O: usize = itr.next().unwrap().parse().unwrap();
    let T: usize = itr.next().unwrap().parse().unwrap();
    let J: usize = itr.next().unwrap().parse().unwrap();
    let L: usize = itr.next().unwrap().parse().unwrap();
    let S: usize = itr.next().unwrap().parse().unwrap();
    let Z: usize = itr.next().unwrap().parse().unwrap();

    let mut ans = 0;

    // use
    let mut tmp = O;
    tmp += I / 2 * 2 + min(J, L) * 2;
    ans = max(ans, tmp);
    // not use
    if I > 0 && J > 0 && L > 0 {
        tmp = O;
        tmp += 3;
        tmp += (I - 1) / 2 * 2 + min(J - 1, L - 1) * 2;
        ans = max(ans, tmp);
    }

    println!("{}", ans);
}