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 C++ (GCC 9.2.1) (judge verdict: Accepted)

/////////// <----------------------  Xorcestor_X ---------------------> ///////////
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
#include <sstream>
#include <algorithm>
#include <map>
#include <iterator>
#include <queue>
#include <utility>
#include <set>
#include <deque>

using namespace std;
#define endl "\n"
#define ll long long int
#define c(x) ll x;cin>>x
#define cc(x,y) ll x,y;cin>>x>>y
#define ccc(x,y,z) ll x,y,z; cin>>x>>y>>z
#define db long double
#define fast cin.tie(NULL); cout.tie(NULL); ios_base::sync_with_stdio(false)
#define inf 200000000000000
#define mod 1000000007
#define mod2 998244353
#define eps 1e-7
#define PI 3.1415926535897932385
#define pb push_back
#define bitc  __builtin_popcountll
#define mp make_pair
#define ff first
#define ss second
#define all(ar) ar.begin(), ar.end()
#define len(x) (ll)(x).size()
#define fr(i,a,b) for (ll i = (a), _b = (b); i <= _b; i++)
#define rep(i,n) for (ll i = 0, _n = (n); i < _n; i++)
#define repr(i,n) for (ll i = n - 1; i >= 0; i--)
#define frr(i,a,b) for (ll i = (a), _b = (b); i >= _b; i--)
#define foreach(it,ar) for (auto it = ar.begin(); it != ar.end(); it++)
#define fill(ar,val) memset(ar, val, sizeof(ar))
#define print(arr,n) cout<<" :: "<<#arr<<" :: "<<endl; rep(i,n) cout<<arr[i]<<" "; cout<<endl
#define print2(arr,n,m) cout<<" :: "<<#arr<<" :: "<<endl; rep(i,n){ rep(j,m) cout<<arr[i][j]<<" "; cout<<endl; }
#define dyn1(arr,n) ll* arr=new ll[n]; rep(i,n) arr[i]=0;
#define dyn2(arr,n,m) ll** arr=new ll*[(n)]; rep(i,n) arr[i]=new ll[m]; rep(i,n) rep(j,m) arr[i][j]=0
#define carr(arr,n) rep(i,n) cin>>arr[i]
#define carr2(arr,n,m) rep(i,n) rep(j,m) cin>>arr[i][j]
#define debug(a) cout<<" :: "<<#a<<" == "<<a<<endl
#define debug2(a,b) cout<<" :: "<<#a<<" == "<<a<<" :: "<<#b<<" == "<<b<<endl
#define watch(x) cerr << (#x) << " is " << (x) << endl
#define debugx(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)
#define time__(d) for(long blockTime = 0; (blockTime == 0 ? (blockTime=clock()) != 0 : false); debugx("%s time : %.4fs", d, (double)(clock() - blockTime) / CLOCKS_PER_SEC))
typedef pair<ll,ll> pii;
typedef vector<pii> vii;

inline ll maxim(ll a,ll b) {if(a>b) return a; else return b;}
inline ll minim(ll a,ll b) {if(a<b) return a; else return b;}
inline bool equals(double a, double b){ return fabs(a - b) < 1e-9; }
ll gcd(ll a, ll b){ return b==0 ? a : gcd(b, a%b); }

ll sum(ll* arr,ll n)
{
  ll total=0;
  rep(o,n)
  {
    total+=arr[o];
  }
  return total;
}

ll Extended_gcd(ll a,ll b,ll& x,ll& y)
{
  if(a==0)
  {
    x=0;
    y=1;
    return b;
  }
  else
  {
    ll x1,y1;
    ll g=Extended_gcd(b%a,a,x1,y1);
    x=y1-(b/a)*x1;
    y=x1;
    return g;
  }
}

bool prime[2000010];
void SieveOfEratosthenes()
{
    memset(prime, true, sizeof(prime));
    prime[1]=false;
    for (int p=2; p*p<=2000001; p++)
    {
        if (prime[p] == true)
        {
            for (int i=p*p; i<=2000001; i += p)
                prime[i] = false;
        }
    }
}


vector<ll> prime_vec[1000010];
void modified_SieveOfEratosthenes()
{
    //prime[1]=false;
    for (int p=2; p*p<=1000001; p++)
    {
        if (prime_vec[p].size()==0)
        {
            for (int i=p*2; i<=1000001; i += p)
                prime_vec[i].pb(p);
        }
    }
}

ll binpow(ll a, ll b)
{
    ll res = 1;
    while (b > 0)
    {
        if (b & 1) res = (res * a);
        a = (a * a);
        b >>= 1;
    }
    return res;
}

ll modpow(ll a, ll b,ll x)
{
    ll res = 1;
    while (b > 0)
    {
        if (b & 1) res = (res * a)%x;
        a = (a * a)%x;
        b >>= 1;
    }
    return res;
}

int modInverse(int a, int m)
{
    int m0 = m;
    int y = 0, x = 1;
    if (m == 1)
      return 0;
    while (a > 1)
    {
        // q is quotient
        int q = a / m;
        int t = m;
        // m is remainder now, process same as
        // Euclid's algo
        m = a % m, a = t;
        t = y;
        // Update y and x
        y = x - q * y;
        x = t;
    }
    // Make x positive
    if (x < 0)
       x += m0;
    return x;
}


ll lpf[1000001] = {};

void precompute()
{
  fr(i,2,1000000)
  {
    if(!lpf[i])
    {
      for(ll j=i;j<=1000000;j+=i)
      {
        if(!lpf[j])
        lpf[j]=i;
      }
    }
  }
}







vector<ll> adj[1000010];
ll degree[1000010];

void adj_former(ll n)
{
  rep(o,n-1)
  {
    cc(a,b);
    adj[a].pb(b);
    adj[b].pb(a);
    degree[a]++;
    degree[b]++;
  }
}

// ll calc(ll idx)
// {
//   ll ret_val=-1;
//   ll x=arr[idx];
//   if(prime[x])
//   {
//     ret_val=maxim(ret_val,p[x]);
//     p[x]=idx;
//     return ret_val;
//   }
//   else
//   {
//     rep(u,prime_vec[x].size())
//     {
//       ret_val=maxim(ret_val,p[prime_vec[x][u]]);
//       p[prime_vec[x][u]]=idx;
//     }
//     return ret_val;
//   }
// }


ll arr[1000010]={0};
ll brr[1000010]={0};
ll crr[1000010]={0};
ll pre[1000010]={0};
ll suff[1000010]={0};
ll freq[1000010];
ll a[1000010]={0};
ll b[1000010]={0};
ll mat[3010][3010];
map<ll,ll> lookup;
string sarr[1000010];
vector<ll> cont;


ll func(ll x)
{
  ll count=0;
  rep(o,26)
  {
    count+=(freq[o]/(x));
  }
  return count;
}


ll bs(ll low,ll high,ll target)
{
  while(low+3<high)
  {
    ll mid=(low+high)/2;
    if(func(mid)<target)
    {
      high=mid-1;
    }
    else
    {
      low=mid;
    }
  }

  frr(o,high,low)
  {
    if(func(o)>=target) return o;
  }
  return -1;//to check for bugs
}


bool leap(ll y)
{
  if(y%400==0) return true;
  if((y%4==0)&&(y%100!=0)) return true;
  return false;
}

ll isprime(ll n)
{
  if(n==1) return false;
  if(n==2) return true;
  bool alive=true;
  for(int i=2;i*i<=n;i++)
  {
    if(n%i==0) return false;
  }
  return true;
}


int main()
{
  fast;

  #ifndef ONLINE_JUDGE
  // for getting input from input.txt
  freopen("input.txt", "r", stdin);
  // for writing output to output.txt
  freopen("output.txt", "w", stdout);
	#endif

  cout<<setprecision(10)<<fixed;
  ll i,o,t,j,l,s,z;
  cin>>i>>o>>t>>j>>l>>s>>z;
  ll total1=(i/2)*2+o+(j/2)*2+(l/2)*2;
  i--;
  j--;
  l--;
  ll total2;
  if(i>=0&&j>=0&&l>=0)
  total2=(i/2)*2+o+(j/2)*2+(l/2)*2+3;
  else total2=0;
  cout<<maxim(total1,total2)<<endl;

  return 0;
}