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;
using ll=long long int;
using ld=long double;
using VI=vector<ll>;
using VD=vector<ld>;
using VVI=vector<VI>;
using VC=vector<char>;
using VB=vector<bool>;
using VVC=vector<VC>;
using VS=vector<string>;
using PLL =pair<ll,ll>;
using PLD=pair<ld,ld>;
using VPLL=vector<PLL>;
#define print(x) std::cout<<x<<"\n"
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
#define repd(i,n) for(ll i=(ll)(n)-1;i>=0;i--)
#define ALL(x) (x).begin(),(x).end()
#define ALLR(x) (x).rbegin(),(x).rend()
#define SZ(x) ((ll)(x).size())
#define MAX(x) *max_element((x).begin(),(x).end())
#define MIN(x) *min_element((x).begin(),(x).end())
#define SORTR(x) sort((x).rbegin(),(x).rend())
#define SORT(x) sort((x).begin(),(x).end())
#define SUM(x) accumulate((x).begin(),(x).end(), 0)
#define FILL(x,a) fill(x.begin(),x.end(),a)
#define EACH(i,x) for(typeof((x).begin()) i=(x).begin(); i!=(x).end(); ++i)
#define EXIST(v, x) (std::find(v.begin(), v.end(), x) != v.end())

const ll INF = 1e18;
const ld EPS   = 1e-10;
const int MOD  = int(1e9)+7;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template <class BidirectionalIterator>
bool next_partial_permutation(BidirectionalIterator first, BidirectionalIterator middle,BidirectionalIterator last){reverse(middle, last); return next_permutation(first , last);}
ll gcd(ll x, ll y) { return (x % y)? gcd(y, x % y): y; }
ll lcm(ll x, ll y) { return x / gcd(x, y) * y; }
ll GCD(VI v){ll a = v[0]; for (ll i = 1; i<SZ(v); i++) {a = gcd(a, v[i]);} return a;}
ll LCM(VI v){ll a = v[0]; for (ll i = 1; i<SZ(v); i++) {a = lcm(a, v[i]);} return a;}
VI Bit2Vector(const ll bit, ll n) {	VI s;	rep(i,n) if (bit & (1 << i)) s.push_back(i); return s;}


void Main()
{
	ll n,m,l; ll res=0;
	string sres="YES";

	ll a,b,v,w,t;

	cin>>a>>v;
	cin >>b>>w;
	cin>>t;
	//距離はaとbの位置の差の絶対値
	ll distance=abs(a-b);
	//速度の差分
	ll diffvelo=v-w;
	//速度がーになるときはいっか
	if(diffvelo<=0) {
		print("NO");
		return ;
	}
	if(distance/t>1||distance/diffvelo>1){
		print("YES");
	}
	res=distance-diffvelo*t;
	if(res>=0) sres="NO";

	cout << sres << "\n";
	return;
}

int main()
{
	std::cin.tie(0);
	std::ios_base::sync_with_stdio(false);
	std::cout << std::fixed << std::setprecision(15);
	Main();
}