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.Scanner;
import java.lang.*;

public class Main{
 
	public static void main(String args[]){

    	//入力するためのオブジェクトを生成
      	Scanner scan = new Scanner(System.in);
	int A = scan.nextInt();	//Aのいる座標
	int V = scan.nextInt();	//Aの秒速
	int B = scan.nextInt(); //Bがいる座標
	int W = scan.nextInt();	//Bの秒速

	int T = scan.nextInt(); //秒数
	


	//AとBの位置関係により進む方向を場合分けする
	if(A < B){ // ABの順でならんでいるとき
	   if(A +( V * T) < B + (W * T) ){
	      System.out.println("No");
	   }else if(A +( V * T) >= B + (W * T) ){
		System.out.println("Yes");
	   }
	// }else if(A +( V * T) >= B + (W * T)){
	  //    System.out.println("Yes");
	   //}else{
	     // System.out.println("No");
	   //}
	 }else if(B < A){ //BAの順で並んでいる時
	   if(B -(W * T) >= A -(V * T) ){
	      System.out.println("Yes");
	   }else if(B -(W * T) <= A -(V * T) ){
	      System.out.println("No");
           }

	
	}

	

   }
}