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)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vl = vector<ll>;
using vvl = vector<vl>;
using pl = pair<ll, ll>;
const ll INF = ll(1e18);
const ll mod = ll(1e9+7);
const double pi = acos(-1);
#define rep0(i,n) for(ll (i) = 0; (i) < (n); ++(i))
#define rrep0(i,n) for(ll (i) = (n) - 1; (i) >= 0; --(i))
#define rep1(i,n) for(ll (i) = 1; (i) <= (n); ++(i))
#define rrep1(i,n) for(ll (i) = (n); (i) >= 1; --(i))
#define nfor(i,a,b) for(ll (i) = (a); (i) < (b); ++(i))
#define pf(x) cout << (x) << endl
#define all(x) (x).begin(),(x).end()
#define yes pf("Yes")
#define no pf("No")
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; }
const int dx[8]={1,0,-1,0,1,1,-1,-1};
const int dy[8]={0,1,0,-1,1,-1,1,-1};

ll gcd(ll a,ll b){
    if(a < b)swap(a, b);
    if(b == 0) return a;
    return gcd(b,a%b);
}

ll lcm(ll a,ll b){
    ll g = gcd(a,b);
    return a / g * b;
}

ll factorial(ll n){
    ll ans = 1;
    rep1(i, n){
        ans *= i;
        ans %= mod;
    }
    return ans;
}

ll power(ll a, ll b){
    ll ans = 1;
    a %= mod;
    while(b) {
        if(b & 1LL) ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans % mod;
}

//modの値の確認をすること
int main(){
    ll a,v,b,w,t;
    cin >> a >> v >> b >> w >> t;
    if((v - w) * t >= abs(b - a))pf("YES");
    else pf("NO");
    return 0;
}