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>
    #define int ll
    #define M 1000000007
    #define hell 998244353
    #define inf 1e18
    #define pb push_back
    #define pi 3.1415926536
    #define mp make_pair
    #define ub upper_bound
    #define lb lower_bound
    #define all(a) a.begin(),a.end()
    #define ll long long
    #define vec vector <int>
    #define pii pair<int,int>
    #define endl "\n"
    #define gcd __gcd
    using namespace std;
    int power(int x, unsigned int y, unsigned int m) 
    { 
        if (y == 0) 
            return 1; 
        int p = power(x, y/2, m) % m; 
        p = (p * p) % m; 
  
        return (y%2 == 0)? p : (x * p) % m; 
    } 
    int modInverse(int a, int m) 
    {   
        return power(a, m-2, m);
    }
    int nCrModPFermat(int n, int r, int p) 
    { 
           if (r==0) 
              return 1; 
  
        int fac[n+1]; 
        fac[0] = 1; 
        for (int i=1 ; i<=n; i++) 
            fac[i] = fac[i-1]*i%p; 
  
        return (fac[n]* modInverse(fac[r], p) % p * 
            modInverse(fac[n-r], p) % p) % p; 
    } 
    signed main()
    {
        ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
        ll t=1;
        // cin >> t;
        /*SAARE INPUTS LENA*/
        /*VARIABLES KI RANGE DEKHLENA*/
        /*EXPONENT RANGE OF OUTPUT KA DHYAN RAKHNA*/
        while(t-->0)
        {
            ll d=0,t=0,s=0;
            cin>>d>>t>>s;
            double x=((double)d)/s;
            if(x<=t)
                cout<<"Yes";
            else
                cout<<"No";
        }
        return 0;
    }