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)

/*input
1 2
3 3
3
*/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
 
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
#define ll long long 
#define ld long double
#define pii pair<int,int>
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(ll i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define MP make_pair
const ll INF64=4e12;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
ll mult(ll a,ll b){
    return ((a%MOD)*(b%MOD))%MOD;
}
ll mypow(ll a,ll b){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1){
            res=(res*a)%MOD;
        }
        a=(a*a)%MOD;
        b>>=1;
    }
    return res;
}
const ll maxn=1e6+5;
const ll maxlg=__lg(maxn)+2;
int main(){
    ios::sync_with_stdio(false),cin.tie(0);
    ll a,v,b,w,t;
    cin>>a>>v>>b>>w>>t;
    if(a!=b and v<=w){
        cout<<"NO";
    }
    else{
        if(a==b) cout<<"YES";
        else{
            if(abs(a-b)<=t*(v-w)){
                cout<<"YES\n";
            }
            else cout<<"NO";
        }
    }
}