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<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
//if(strcmp(s[i],"AC")==0)
//if(s[l]=='#')
/*小文字判定
if((moji>= 'a') && (moji<='z')){
    printf("a\n");
    return 0;
  }
  */
//文字列中の一文字⇒数字の時は s[i]-'0'
int W_Length(char *str){
	int length = 0;
	while(*str++ != '\0')
		length++;
	return length;
}

void swap(long long int *a, long long int *b){
  int temp;
  temp=*a;
  *a=*b;
  *b=temp;
}
//swap(&a, &b)

double deg(double theta){//deg to theta
return theta*(2*3.14159265359)/360;
}

long long int cmp( const void *p, const void *q ) {
    return *(long long int*)p - *(long long int*)q;
}//ちいさい順
//qsort(a,n,sizeof(long long int),cmp);


int cmp_s(const void * left, const void * right)
{
	char *left_char = (char *)left;
  char *right_char = (char *)right;
  return strcmp(left_char, right_char);
}
//qsort(s,n,sizeof(s[0]),cmp_s);
//char s[n][11]

long long int GCD(long long int a, long long int b){
  long long int L=a,S=b;
  while(1){
    if(S>L){
      L=S+L;
      S=L-S;
      L=L-S;
    }
    if(L%S==0){
      return S;
    }
    L=L%S;
  }
  
}

long long int LCM(long long int a, long long int b){
  return a*b/(GCD(a,b));
}

int main (void)
{
	long long int n,i,min,d,t,s,k,a[210000],max,j=1,count=0;
scanf("%lld%lld%lld",&d,&t,&s);
if(s*t>=d){
  printf("Yes");
  return 0;
}
  
  printf("No");
return 0;
}