CF#256(Div.2) 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 and a3third 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 1
1 1 1
4

output
YES

input
1 1 3
2 3 4
2

output
YES

input
1 0 0
1 0 0
1

output
NO  题意就是Bizon the Champion这个人得了非常多奖，有a1个一等奖奖杯。a2个二等奖奖杯。a3个三等奖奖杯。b1张一等奖奖状。b2张二等奖奖状，b3张三等奖奖状。如今给你n个柜子。问你能否将这些奖杯和奖状放下，规则是奖杯和奖状不能放在同一个柜子里。一个柜子最多仅仅能放5个奖杯或10张奖状。
解题思路：将现有的奖杯和奖状所须要的柜子书求出。假设小于n，则输出“YES”;否则输出“NO”。#include<stdio.h>
int s[3],y[3];
int main()
{
    int b,d,m;
    scanf("%d %d %d",&s[0],&s[1],&s[2]);
    int a=s[0]+s[1]+s[2]+4;//一个小技巧，加上4以后能够将不足5个所需的柜子书加上！

	b=a/5;
    scanf("%d %d %d",&y[0],&y[1],&y[2]);
    int c=y[0]+y[1]+y[2]+9;//同上
	d=b/10;
    scanf("%d",&m);
    if(b+d>m)
        printf("NO
");
    else
        printf("YES
");
    return 0;
}