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
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
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.
題意：有個人想要去旅行，他的貓吃東西比較挑剔，
週一，週四，週日 吃魚
週二，週六吃rabbit stew
其他的，也就是週三，週五吃chicken stake
輸入三中食品帶的量，求最多能吃幾天。
解法：首先，我們求出來a/3,b/2,c/2的最小值，是我們的結果，a,b,c減去這幾周所需要消耗的，因爲可能是週日，週一週二。。。這種情況，所以我們兩層for取餘7，判斷就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;
}