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Dorchester Center, MA 02124

Eikooc loves bread. She has *N* loaves of bread, all of which expire after exactly *M* days. She can eat upto *K* loaves of bread in a day. Can she eat all the loaves of bread before they expire?

- The first line contains a single integer
*T*– the number of test cases. Then the test cases follow. - Each test case consists of a single line containing three integers
*N*,*M*and*K*– the number of loaves of bread Eikooc has, the number of days after which all the breads will expire and the number of loaves of bread Eikooc can eat in a day.

For each test case, output `Yes`

if it will be possible for Eikooc to eat all the loaves of bread before they expire. Otherwise output `No`

.

You may print each character of `Yes`

and `No`

in uppercase or lowercase (for example, `yes`

, `yEs`

, `YES`

will be considered identical).

- 1≤
*T*≤1000 - 1≤
*N*,*M*,*K*≤100

**Input**: 3
100 100 1
9 2 5
19 6 3
**Output**: Yes
Yes
No

**Test case 1:** Eikooc can eat one loaf of bread per day for 100 days. Thus none of the bread expires.

**Test case 2:** Eikooc can eat 5 loaves of the first day and 4 loaves on the second day. Thus none of the bread expires.

**Test case 3:** There is no way Eikooc can consume all the loaves of bread before it expires.

```
import java.util.*;
import java.lang.*;
import java.io.*;
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int a=sc.nextInt();
int b=sc.nextInt();
int c=sc.nextInt();
if(a<=b*c)
System.out.println("yes");
else
System.out.println("no");
}
}
}
```

```
#include <iostream>
using namespace std;
int main() {
int t;
cin>>t;
while(t--){
int n, m, k;
cin>>n>>m>>k;
if((m*k)>=n){
cout<<"Yes"<<endl;
}
else{
cout<<"No"<<endl;
}
}
return 0;
}
```

```
t=int(input())
for i in range(t):
a,b,c=map(int,input().split())
if b*c>=a:
print("YES")
else:
print("NO")
```

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