Codeforces Round #552 (Div. 3) C. Gourmet Cat

Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:

on Mondays, Thursdays and Sundays he eats fish food;
on Tuesdays and Saturdays he eats rabbit stew;
on other days of week he eats chicken stake.

Polycarp plans to go on a trip and already packed his backpack. His backpack contains:

aa daily rations of fish food;
bb daily rations of rabbit stew;
cc daily rations of chicken stakes.

Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.

Input

The first line of the input contains three positive integers aa, bb and cc (1≤a,b,c≤7⋅1081≤a,b,c≤7⋅108) — the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.

Output

Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.

Examples

Input

2 1 1

Output

Input

3 2 2

Output

Input

1 100 1

Output

Input

30 20 10

Output

Note

In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday — rabbit stew and during Wednesday — chicken stake. So, after four days of the trip all food will be eaten.

In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.

In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 9999 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip.

题意：有个人想要去旅行，他的猫吃东西比较挑剔，

周一，周四，周日吃鱼

周二，周六吃rabbit stew

其他的，也就是周三，周五吃chicken stake

输入三中食品带的量，求最多能吃几天。

解法：首先，我们求出来a/3,b/2,c/2的最小值，是我们的结果，a,b,c减去这几周所需要消耗的，因为可能是周日，周一周二。。。这种情况，所以我们两层for取余7，判断就ok了

#include<bits/stdc++.h>
using namespace std;
int main(){
  int a , b , c , ans  , n ;
  while(~scanf("%d%d%d",&a,&b,&c)){
    int mi = min(a/3,min(b/2,c/2));
    // cout<<mi<<endl;
    ans = mi * 7;
    a -= mi * 3;
    b -= mi * 2;
    c -= mi * 2;
    n = -1;
    // cout<<a<<"+++++"<<b<<" "<<c<<endl;
    for(int i = 1 ; i <= 7 ; i ++){
      int cnt = 0;
      int a1 = a ;
      int b1 = b;
      int c1 = c;
      // cout<<a1<<"   "<<b1<<"    "<<c1<<endl;
      for(int j = 0 ; j <= 6 ; j ++){
        int t = (i + j) % 7;
        if(t == 1 || t == 4 || t == 0){
          if(a1){
            a1 --;
            cnt ++;
          }
          else
            break;
        }
        else if(t == 2 || t == 6){
          if(b1){
            b1 --;
            cnt ++;
          }
          else
            break;
        }
        else if(t == 3 | t == 5){
          if(c1){
            c1 -- ;
            cnt ++;
          }
          else
            break;
        }
      }
      n = max(n,cnt);
    }
    // cout<<n<<endl;
    cout<<ans+n<<endl;
  }
  return 0;
}