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: Accepted)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define rep2(i,n) for(int i=0;i<=n;i++)
#define repr(i,a,n) for(int i=a;i<n;i++)
#define all(a) a.begin(),a.end()
#define P pair<long long,long long>
#define uni(e) e.erase(unique(e.begin(),e.end()),e.end())
#define double long double
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
int INF=1e10;
int MOD=1e9+7;
template<class T> void out(T a){cout<<a<<'\n';}
template<class T> void yesno(T b){if(b)out("yes");else out("no");}
template<class T> void YesNo(T b){if(b)out("Yes");else out("No");}
template<class T> void YESNO(T b){if(b)out("YES");else out("NO");}
template<class T> void noyes(T b){if(b)out("no");else out("yes");}
template<class T> void NoYes(T b){if(b)out("No");else out("Yes");}
template<class T> void NOYES(T b){if(b)out("NO");else out("YES");}
int keta(int a){
  double b=a;
  b=log10(b);
  int c=b;
  return c+1;
}
int kurai(int a,int b){
  int m=pow(10,b),n=pow(10,b-1);
  return (a%m)/n;
}
int ketawa(int a){
  int k=0;
  rep(i,keta(a)){
  k+=kurai(a,i+1);
  }
  return k;
}
int gcd(int a,int b){
  if(a%b==0)
    return b;
  return gcd(b,a%b);
}
int lcm(int a,int b){
    return a/gcd(a,b)*b;
}
int mid(int a,int b,int c){
  vector<int> k(3);
  k[0]=a;k[1]=b;k[2]=c;
  sort(all(k));
  return k[1];
}
bool sosuu(int a){
  bool b=1;
  if(a<=1)
    return 0;
  else{
  rep1(i,sqrt(a)-1){
  if(a%(i+1)==0)
  b=0;
  }
  return b;
  }
}
int modpow(int a,int b,int m=MOD){
 if(!b)return 1; if(b&1)return modpow(a,b-1,m)*a%m;
 int memo = modpow(a,b>>1,m);
 return memo*memo%m;
}
int fact_mod(int n) {
    int f=1; 
for(int i=2;i<n+1;i++) f=f*(i%MOD)% MOD;
    return f;
}
int mod_pow(int x,int n) {
    int res=1;
    while(n>0){
        if(n&1) res=(res*x)%MOD; 
        x=(x*x)%MOD;        
n>>=1; 
    }
    return res;
}
int c_mod(int n, int r) {
    if(r>n-r) r=n-r;
    if(r==0) return 1;
    int a=1;
    rep(i,r) 
a=a*((n-i)%MOD)%MOD;
    int b=mod_pow(fact_mod(r), MOD-2);
    return (a%MOD)*(b%MOD)%MOD;
}
signed main(){
  double a,b,c,d,e;
  cin>>a>>b>>c>>d>>e;
  if(b-d!=0)
  YESNO(e>=(abs(c-a)/(b-d))&&(abs(c-a)/(b-d))>0);
  else
    out("NO");

}