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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;


void init(){
    cin.tie(0);
    cin.sync_with_stdio(0);
}

#define ll long long
#define ld long double
#define ull unsigned long long
#define ull unsigned long long
#define mod 1000000007
#define pi (3.1415926)

#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))

#define scd(a)  scanf("%lld",&a)
#define scdd(a,b)	scanf("%lld%lld",&a,&b)
#define scddd(a,b,c)	scanf("%lld%lld%lld",&a,&b,&c)
#define ptab(a)	printf("%lld ",a)
#define pdn(a)	printf("%lld\n",a)
#define pddn(a,b)	printf("%lld %lld\n",a,b)
#define pdddn(a,b,c)	printf("%lld %lld %lld\n",a,b,c)

#define Filein(N) freopen("N", "r", stdin);
#define Fileout(N) freopen("N", "w", stdout);

#define forr(n) for(ll i=0;i<n;i++)
#define rev(n) for(ll i=n-1;i>=0;i--)
#define forp(x, y) for(ll i=x;i<y;i++)
#define loop( j, k, in) for (ll i=j ; i<k ; i+=in)

#define en "\n"
#define sp " "
#define eps 1.0E-14
#define l length()
#define preci(x) cout << fixed<<setprecision(x)
#define inf LONG_LONG_MAX
#define ninf LONG_LONG_MIN

ull fastpow(ull a,ull b)
{
  ull wynik=1;
  while (b)
  {
    if (b&1)wynik=(wynik*a)%mod;
    a=(a*a)%mod;
    b>>=1;
  }
  return wynik;
}

bool odd(int n)
    {
        int x = sqrt(n);
    for (int i=2 ; i<=x ; ++i)

    {
        if(n%i == 0)return false;
    }

    return true;
    }
ll fact(int n ){

long long fact =1;

for (int i =2 ;i<=n;++i)

{
    fact*=i;
}
return fact;
}

bool prim(long long a){

long long s = sqrt(a);

for (int i=2 ; i<= s ;++i){

    if (a % i == 0)return false;
}
return true;
}

ll sum (int n ){return ((n*(n+1)) / 2) ;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

bool myfunction (string i,string j) /* this function to check who is grater in
sorting strings as numbers ////////
int n;
    cin>>n;
    vector <string> arr(n);
    for(int i=0;i<n;i++)
        cin>>arr[i];


    sort(arr.begin(),arr.end(),myfunction);

    for(int i=0;i<n;i++)
        cout<<arr[i]<<endl;
*/
{
    int n=i.length();
    int m=j.length();
    if(n==m)
        return (i<j);

    return n<m;
}
long long solve (int a, int b)
{
    return min(a,min(b , (a+b) / 3));
}

int main()
{
    init();
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    long long a, b, c,d , e; cin>>a>>b>>c>>d>>e;

    if ((ll) abs(a-d) <= (b-d)*e) cout << "YES";
    else cout << "NO";
}