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 C (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <math.h>

int main(void)
{
	int A=0,B=0,V=0,W=0,T=0; 
	double A1=0,B1=0,V1=0,W1=0,T1=0; 
	int disA=0,disB=0;
	//printf("狼位置Aと1s当たりに進む距離Vを入力してください\n");
	scanf("%d %d", &A,&V);
	//scanf("%d", &B);
	//printf("子羊位置Bと1s当たりに進む距離Wを入力してください\n");
	scanf("%d %d", &B, &W);
	//scanf("%d", &W);
	//printf("鬼ごっこをする時間Tを入力してください\n");
	scanf("%d", &T);

	//数値確認
	A1 = (pow(10, 9.0))*(-1);
	B1 = (pow(10, 9.0));
	V1 = 1;
	W1 = (pow(10, 9.0));
	T1 = (pow(10, 9.0));

	if(A1 <= A && B <= B1){
		if(V1 <= V && W <= W1){
			if(1<=T && T<=T1){
				if(A!=B){
					disA = (A+V*T);
					disB = (B+W*T);
					if(disA >= disB){
						printf("YES\n");
					}else{
						printf("NO\n");
					}
				}
			}else{
				return 0;
			}
		}else{
			return 0;
		}

	}else{
		return 0;
	}

	return 0;
}