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 <iostream> // cout, endl, cin
#include <string> // string, to_string, stoi
#include <cstring>
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset> // bitset
#include <cctype> // isupper, islower, isdigit, toupper, tolower
#include <cmath>
#include <numeric>
#include <float.h>
#include <math.h>
#include <sstream>
#include <iterator>
using namespace std;
// repマクロの定義
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
// allマクロの定義
#define all(v) v.begin(), v.end()
// 大きい順
#define rall(v) v.rbegin(), v.rend()
#define vi vector<int>
#define vl vector<ll>
const  long long INF = 1LL << 60;
typedef  long long ll;
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}



int ctoi(char c) {
 	switch (c) {
		case '0': return 0;
		case '1': return 1;
		case '2': return 2;
		case '3': return 3;
		case '4': return 4;
		case '5': return 5;
		case '6': return 6;
		case '7': return 7;
		case '8': return 8;
		case '9': return 9;
		default: return 0;
	}
}

bool is_prime(const unsigned n){
    switch(n){
        case 0: // fall-through
        case 1: return false;
        case 2: // fall-through
        case 3: return true;
    } // n > 3 が保証された

    if(n % 2 == 0 || n % 3 == 0) return false;
    // n は 2 と 3 のいずれの倍数でもないことが保証された
    // これより n は (6の倍数)-1 か (6の倍数)+1 である

    // 6の倍数前後の数を使って試し割りをする
    for(unsigned i = 5; i * i <= n; i += 6) {
        if(n % i     == 0) return false; // (6の倍数)-1
        if(n % (i+2) == 0) return false; // (6の倍数)+1
    }

    return true;
}

map<ll,ll> mp;

ll primeFactorization(ll a){ //素因数分解を出力するプログラム
    ll i,sq;
    if(a%2==0){ //偶数の場合
        if(!mp.count(2)){
            mp[2] = 1;
        }
        else{
            mp[2] +=1;
        }
        return primeFactorization(a/2); //2で割った値で再帰
    }
    sq = sqrt(a);
    for(i=3;i<=sq;i+=2){ //3以上√a以下の奇数の場合
        if(a%i==0){
            if(!mp.count(i)){
                mp[i] = 1;
            }
            else{
                mp[i] += 1;
            }
            return primeFactorization(a/i); //割れた値で再帰
        }
    }
    //偶数でも3以上√a以下の奇数の場合でも割り切れない場合
    if(a!=1){ //aが1でないなら、a自身は素数
        if(!mp.count(a))
        mp[a] = 1;
        else{
            mp[a] += 1;
        }
    }
    return 0;
}


ll kaizyou(ll n){
    if(n == 1){
        return 1;
    }
    else{
        return (n * kaizyou(n-1)) % 1000000007;
    }
}



int main(){
ll A,V,B,W,T; cin >> A >> V >> B >> W >> T;
ll d;
d = B - A;
ll v = V - W ;
if(d < 0){
        cout << "NO" << endl;
        return 0;
}
else{
    if(v <= 0) {
        cout << "NO" << endl;
        return 0;
    }
    else if(d / v <= T && d % v == 0){
        cout << "YES" << endl;
    }
    else cout << "NO" << endl;
}
}