Codeforces Round #256 (Div. 2) A Rewards

A. Rewards

Bizon the Champion is called the Champion for a reason.

Bizon the Champion has recently got a present — a new glass cupboard with n shelves and he decided to put all his presents there. All the presents can be divided into two types: medals and cups. Bizon the Champion has a₁ first prize cups, a₂ second prize cups anda₃ third prize cups. Besides, he has b₁ first prize medals, b₂ second prize medals and b₃ third prize medals.

Naturally, the rewards in the cupboard must look good, that‘s why Bizon the Champion decided to follow the rules:

• any shelf cannot contain both cups and medals at the same time;
• no shelf can contain more than five cups;
• no shelf can have more than ten medals.

Help Bizon the Champion find out if we can put all the rewards so that all the conditions are fulfilled.

Input

The first line contains integers a₁, a₂ and a₃ (0 ≤ a₁, a₂, a₃ ≤ 100). The second line contains integers b₁, b₂ and b₃ (0 ≤ b₁, b₂, b₃ ≤ 100). The third line contains integer n (1 ≤ n ≤ 100).

The numbers in the lines are separated by single spaces.

Output

Print "YES" (without the quotes) if all the rewards can be put on the shelves in the described manner. Otherwise, print "NO" (without the quotes).

Sample test(s)

Input

1 1 11 1 14

Output

YES

Input

1 1 32 3 42

Output

YES

Input

1 0 01 0 01

Output

NO

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
    int a1,a2,a3,b1,b2,b3,n;
    while(scanf("%d%d%d%d%d%d",&a1,&a2,&a3,&b1,&b2,&b3)!=-1)
    {
        cin>>n;
        int a,b,sum;
        sum=0;
        a=a1+a2+a3;
        b=b1+b2+b3;
        if(a%5==0)
        sum+=a/5;
        else
        sum+=a/5+1;
        if(b%10==0)
        sum+=b/10;
        else
        sum+=b/10+1;
        if(sum<=n)
        cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    return 0;
}