Problem:
Score : 100 points

Problem Statement

Takahashi is meeting up with Aoki.

They have planned to meet at a place that is D meters away from Takahashi's house in T minutes from now.

Takahashi will leave his house now and go straight to the place at a speed of S meters per minute.

Will he arrive in time?

Constraints

1 \leq D \leq 10000

1 \leq T \leq 10000

1 \leq S \leq 10000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

D T S

Output

If Takahashi will reach the place in time, print Yes; otherwise, print No.

Sample Input 1

1000 15 80

Sample Output 1

Yes

It takes 12.5 minutes to go 1000 meters to the place at a speed of 80 meters per minute. They have planned to meet in 15 minutes so he will arrive in time.

Sample Input 2

2000 20 100

Sample Output 2

Yes

It takes 20 minutes to go 2000 meters to the place at a speed of 100 meters per minute. They have planned to meet in 20 minutes so he will arrive just on time.

Sample Input 3

10000 1 1

Sample Output 3

No

He will be late.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
//#include<iostream>


#define ll          long long
#define uint        unsigned int 
#define ld          long double
#define SAM         ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define FOR(i,a,b)  for(int i=a;i<b;++i)
#define RFOR(i,a,b) for(int i=a;i>b;--i)
#define mod         1000000007
#define mxn         200005
#define pb          push_back
#define pii         pair<int,int>
#define f           first
#define s           second
#define sz(x)       (int)x.size()
#define all(x)      x.begin(),x.end()

using namespace std;

/*mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int power(int a,int n)
{
    a %= mod;
    if(n == 1) return a;
    if(n == 0) return 1;
    if(n%2) return (a*(power((a*a)%mod,n/2)%mod))%mod;           //n is odd
    return power((a*a)%mod,n/2)%mod;                             //n is even
}

const int inf = (int) 1e18;

int inverse(int x)
{
    return power(x,mod-2)%mod;     //Fermat's little theorem
}*/

int main()
{
    SAM;
    int d,t,s;
    cin>>d>>t>>s;
    
    ll temp=t*s;
    if(temp>=d)
        cout<<"Yes";
    else
        cout<<"No";
   
   
    
    return 0;
}