Impossible is nothing

Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

Problem Description

Nanae is a boy who loves a girl named Amo,but Amo says that it is unrealistic because they are far away from each other.In order to show his determination,Nanae decides to see Amo by passing through a road where he will meet many kinds of difficult.

Now Nanae,who has zero RP and zero HG initially,is standing in front of the road as the picture shown,and when he passes through one part of the road(one of (a)、(b)、(c)、(d) as in the picture),his RP will be calculated by the rule as shown in the picture,and his HG will increase by one in the meantime.Each part of the road can be passed repeatedly.

Amo is considering to test Nanae,so she tells Nanae that if he can get at least 100 RP and at most n HG when he leaves the maze,she will be willing to be his girlfriend.But be attention,Nanae can only pass through any part of the road with the same direction as his first passing of this part.(for example,if Nanae passes through (b) from left to right,he can only choose this direction(from left to right) if he passes (b) again.)

Now suppose you are Nanae,and the "barrier number" of (a)、(b)、(c)、(d) will be changed.Given the four numbers of (a)、(b)、(c)、(d) and the upper bound of HG,please draw a conclusion that if Amo will become Nanae's girlfriend.


The input contains several test cases.

The first line contains four integers a, b, c, d, which stand for the numbers in the maze.

The second line contains n, which stands for the upper bound of HG.

0 <= a, c, d <= 1000

0 < b, n <= 1000

note that "x ÷ b" = int(x/b)


For each test case,output "Yes" if Amo will become Nanae's girlfriend,and "No" for the opposite condition.

Sample Input

5 5 5 5
1 1 1 1

Sample Output





Solved Number13
Submit Number78
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