Problem:
Score : 200 points

Problem Statement

Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second.
The other child is now at coordinate B, and she can travel the distance of W per second.

He can catch her when his coordinate is the same as hers.
Determine whether he can catch her within T seconds (including exactly T seconds later).
We assume that both children move optimally.

Constraints

-10^9 \leq A,B \leq 10^9

1 \leq V,W \leq 10^9

1 \leq T \leq 10^9

A \neq B

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A V
B W
T

Output

If "it" can catch the other child, print YES; otherwise, print NO.

Sample Input 1

1 2
3 1
3

Sample Output 1

YES

Sample Input 2

1 2
3 2
3

Sample Output 2

NO

Sample Input 3

1 2
3 3
3

Sample Output 3

NO

Solution in Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)


import java.util.Map.Entry;
import java.util.*;
import java.math.*;
import org.w3c.dom.css.Counter;
  
public class Main{
    public static void main(final String[] args){
 
        final Scanner sc = new Scanner(System.in);
        int A =sc.nextInt();
        int V =sc.nextInt();
        int B =sc.nextInt();
        int W =sc.nextInt();
        int T =sc.nextInt();
        int div = Math.abs(A-B);
        int run = V-W;
        int ans =div -run*T;
        if(ans>=0){
            System.out.println("NO");
        }else{
            System.out.println("YES");
        }

        
        /* 
        long N = sc.nextLong();
        int ans=0;
        long q =2;
 
        //1だった時はさっさと終わらす
        if(N==1){
            System.out.println(0);
            return;
        }
 
        //本編
        while(q*q<=N){
            //qを決定する
            if(N%q ==0){
                //qの累積を全部発見する
                for(int i = 1; N>=Math.pow(q,i);i++){
                    if(N%Math.pow(q, i)==0){
                        //発見するたびにカウントアップ
                        ans++;
                        //発見するたびに除算
                        N = (long) (N / Math.pow(q, i));
                    }
                }
                //qの累積は見つけ尽くしたので、Nからqを完全に取り除く
                while(N%q ==0){
                    N = N/q;
                }
            }   
            q++;
        }

        //1以外の数が搾りカスとして出たら足してあげる
        if(N != 1){ans++;}
            System.out.println(ans);*/
    }
}