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)

//
//東京海上日動　プログラミングコンテスト2020
//B - Tag
//

#include <iostream>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <queue>
#include <map>
#include <set>
#include <tuple>
#include <functional>
#include <cassert>

using namespace std;

#define REP(i,n) for(ll i=0;i<(ll)(n);i++)
#define REPD(i,n) for(ll i=n-1;i>=0;i--)
#define FOR(i,a,b) for(ll i=a;i<=(ll)(b);i++)
#define FORD(i,a,b) for(ll i=a;i>=(ll)(b);i--)
#define ALL(x) (x).begin(),(x).end()
#define RALL(x) (x).rbegin(),(x).rend()
#define SIZE(x) (int)(x).size()
#define SIZEL(x) (ll)(x).size()
#define MSG(a) cout << #a << " " << a << endl;

using ll = long long;
using ld = long double;

const double PI = 3.14159265358979323846;

ll gcd(ll x, ll y) { return (x % y)? gcd(y, x % y): y; }    //最大公約数
ll lcm(ll x, ll y) { return x / gcd(x, y) * y; }            //最小公倍数

int main()
{
    ll a,v,b,w,t;
    cin >> a >> v;
    cin >> b >> w;
    cin >> t;

    string ans = "NO";
    if (v<=w) {
        cout << ans << endl;
        return 0;
    }

    if(abs(b-a)%(v-w) == 0){
        if (1<=abs(b-a)/(v-w) && abs(b-a)/(v-w)<=t) ans="YES";
    }

/*
    if (a<b && v>w){
        ll ta = (b-a)/(v-w);
        if ((b-a)%(v-w) == 0 && ta>0) {
            if (ta<=t) ans = "YES";
        }
    }else if (a>b && v<w){
        ll ta = (a-b)/(w-v);
        if ((a-b)%(w-v) == 0 && ta>0) {
            if (ta<=t) ans = "YES";
        }
    }
*/
    cout << ans << endl;
    return 0;
}