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;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vint;
typedef vector<double> vd;
typedef vector<ll> vll;
typedef vector<string> vstr;
typedef vector<vector<int>> vvint;
typedef vector<pair<int, int>> vpii;
typedef vector<pair<ll, ll>> vpll;
typedef priority_queue<int, vector<int>, greater<int>> spqint; //小さい順に取り出し
typedef priority_queue<ll, vector<ll>, greater<ll>> spqll;     //小さい順に取り出し
typedef priority_queue<int, vector<int>, less<int>> bpqint;    //大きい順に取り出し
typedef priority_queue<ll, vector<ll>, less<ll>> bpqll;        //大きい順に取り出し

#define REP(i, n) for (ll i = 0; i < (ll)(n); i++)
#define FOR(i, a, b) for (ll i = (ll)a; i < (ll)b; i++)
#define IREP(i, v) for (auto i = (v).begin(); i != (v).end(); i++)
#define FI first
#define SE second
#define MP make_pair
#define MT make_tuple
#define PB push_back
#define PF push_front
#define TS to_string
#define BS binary_search
#define LB lower_bound
#define UB upper_bound
#define NP next_permutation
#define ALL(v) (v).begin(), (v).end()
#define SZ(x) (ll) x.size()
#define SP(x) setprecision((ll)x)


int INF = 1e9;
int NIL = -1;
ll MOD = 1000000007;
ll LINF = 1e18;
double EPS = 1e-9;
double PI = M_PI;
int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; } //最大公約数
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }     //最小公倍数

void yes(){ cout << "YES" << endl;}
void no(){ cout << "NO" << endl;}
//-----------------------------------------

//-----------------------------------------
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    std::chrono::system_clock::time_point start,end;
    start = std::chrono::system_clock::now();

    ll a, v ;
    cin >> a >> v ;
    ll b, w ;
    cin >> b >> w ;
    ll t ;
    cin >> t ;
    if(a<b){
        if(a+v*t>=b+w*t){
            yes() ;
        }
        else{
            no() ;
        }
    }
    else{
        if(b-w*t<=a-v*t){
            yes() ;
        }
        else{
            no() ;
        }
    }



    end = std::chrono::system_clock::now();
    auto elapsed = std::chrono::duration_cast< std::chrono::milliseconds >(end - start).count();
    //std::cout << elapsed <<"ms"<< std::endl;

    

    return 0;
}