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>
#include <stdio.h>

using namespace std;
#define inputInt(a) int a; cin >> a; 
#define inputInt2(a,b) int a; int b; cin >> a >> b; 
#define inputInt3(a,b,c) int a; int b; int c; cin >> a >> b >> c;
#define inputLong(a) long a;cin >> a; 
#define inputIntArray(a,N) int a[N];for(int i=0;i<N;i++){cin >> a[i];}
#define inputLongArray(a,N) long a[N];for(int i=0;i<N;i++){cin >> a[i];}
#define inputIntArray2(a,b,N) int a[N]; int b[N]; for(int i=0;i<N;i++){cin >> a[i] >> b[i];}
#define output(answer) cout << answer << endl;
#define fN(i,N)    for(int i=0; i<N; i++)
#define fSE(i,s,e) for(int i=s; i<=e; i++)
#define ll long long
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define FOR(i,N) for(int i=0; i<N; i++)
#define cinv(v,N) vector<int> v(N); for(int i=0; i<N; i++){ cin >> v[i]; }
#define mt make_tuple
#define vll vector<long long int>
#define vvll vector<vector<long long int>>
#define all(v) (v).begin(),(v).end()
#define show(v) FOR(i,v.size()){ cout << v[i] << " "; } cout << endl;
#define showshow(v) FOR(i,v.size()){ FOR(j,v[i].size()){ cout << v[i][j] << " "; } cout << endl; }
const ll MOD = 1e9+7;
const ll MOD2 = 998244353;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

//Combination計算
const int LIMIT = 1e6;//1e6→50ms,1e7→500ms,1e8→5000ms程度。
//const ll MOD = 1e9+7;
vector<ll> FacMod(LIMIT), InvMod(LIMIT), FacInvMod(LIMIT);
const int dpComTableLIMIT = 3e3+1;
vector<vector<ll>> Com(dpComTableLIMIT,vector<ll>(dpComTableLIMIT,0));
const int ComKTableLIMIT = 1e7;
vector<ll> ComK(ComKTableLIMIT);

void DPComInit(){
  //Com[0][j]は0で初期化されている。
  for(int i=0; i<dpComTableLIMIT; i++){ Com[i][0] = 1; }
  
  for(int i=1; i<dpComTableLIMIT; i++){
    for(int j=1; j<dpComTableLIMIT; j++){
      Com[i][j] = (Com[i-1][j-1]+Com[i-1][j])%MOD;
    }
  }
}

void ComKInit(ll N){
  for(int i=0; i<=ComKTableLIMIT; i++){
    if(0<i && i<=N/2){
      ComK[i] = (ComK[i-1]*(N-i+1)/i)%MOD;
    }else if(i==0){
      ComK[i] = 1;
    }else if(N/2<i && i<=N){
      ComK[i] = ComK[N-i];
    }else{//i>Nの時
      ComK[i] = -1;
    }
  }
}

void ComInit(){//tableを作成
  FacMod[0] = FacMod[1] = 1;
  InvMod[0] = 0; InvMod[1] = 1; //InvMod[0]は定義できない
  FacInvMod[0] = FacInvMod[1] = 1;
  
  for(int i=2; i<LIMIT; i++){
    FacMod[i] = FacMod[i-1]*i%MOD;
    InvMod[i] = -InvMod[MOD%i]*(MOD/i)%MOD + MOD;
    FacInvMod[i] = FacInvMod[i-1] * InvMod[i]%MOD;
  }
}

ll ComMod(int n, int k){
  if(n>=0 && k>=0 && n>=k){
    return FacMod[n]*(FacInvMod[k]*FacInvMod[n-k]%MOD)%MOD;
  }else{
    return 0; //戻り値を0にする方が計算結果に影響が少ない。
  }
}


//繰り返し2乗法
ll power(ll x, ll y){//繰り返し2乗法
  if(y < 0){
    return 0;
  }else if(y == 0){
    return 1LL;
  }else if(y%2 == 0){//偶数
    return power(x,y/2)*power(x,y/2);
  }else{//奇数
    return power(x,y-1)*x;
  }
}

//繰り返し2乗法
ll mpower(ll x, ll y, ll m){//繰り返し2乗法
  if(y < 0){
    return 0;
  }else if(y == 0){
    return 1LL;
  }else if(y%2 == 0){//偶数
    return mpower(x,y/2,m)*mpower(x,y/2,m)%m;
  }else{//奇数
    return mpower(x,y-1,m)*x%m;
  }
}

//pair型のソート
bool compare_by_second(pair<ll,ll> a, pair<ll,ll> b){
  return a.second != b.second ? a.second < b.second : a.first < b.first;
}

class UnionFind {
  vector<int> parent;
  vector<int> rank;
  
  public:
  
  UnionFind(int N);
  int root(int x);
  void unite(int x, int y);
  bool isUnited(int x, int y);
  int getRank(int x);
};

UnionFind::UnionFind(int N):parent(N),rank(N,1){
  for(int i=0; i<N; i++){parent[i]=i;}
}
  
int UnionFind::root(int x){
  if(parent[x] == x) return x;
  return root(parent[x]);
  /*parent[x] = parent[parent[x]];
  if(parent[x]!=parent[parent[x]]) rank[x]-=1;*/
}
  
void UnionFind::unite(int x, int y) { // xとyの木を併合
  int rx = root(x); //xの根をrx
  int ry = root(y); //yの根をry
  if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
  
  if(rank[x] >= rank[y]){parent[ry] = rx; rank[y]+=1;}//大きい木に小さい木が併合される
  else{parent[rx] = ry; rank[x]+=1;}//根に直接辺を張る
}
  
bool UnionFind::isUnited(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
  int rx = root(x);
  int ry = root(y);
  parent[x]=rx; parent[y]=ry;rank[x]=2;rank[y]=2;
  return rx == ry;
}

int UnionFind::getRank(int x){
  return rank[x];
}

int main()
{
  //input
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  ll a, v, b, w, t; cin >> a >> v >> b >> w >> t;
  string ans = "YES";
  if(v<w){
    ans = "NO"; 
  }else if(abs(a-b) > (v-w)*t){
    ans = "NO"; 
  }
  
  // output
  cout << ans << endl;

  return 0;
}