**Physical Address**

304 North Cardinal St.

Dorchester Center, MA 02124

Janmansh has to submit 33 assignments for Chingari before 1010 pm and he starts to do the assignments at X*X* pm. Each assignment takes him 11 hour to complete. Can you tell whether he’ll be able to complete all assignments on time or not?

- The first line will contain T
*T*– the number of test cases. Then the test cases follow. - The first and only line of each test case contains one integer X
*X*– the time when Janmansh starts doing the assignemnts.

For each test case, output `Yes`

if he can complete the assignments on time. 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 \le T \le 101≤
*T*≤10 - 1 \le X \le 91≤
*X*≤9

```
Input: 2
7
9
Output: Yes
No
```

**Test case-1:** He can start at 77pm and finish by 1010 pm. Therefore he can complete the assignments.

**Test case-2:** He can not complete all the 33 assignments if he starts at 99 pm.

```
/* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while (t-- > 0) {
int x = Integer.parseInt(br.readLine());
System.out.println(10 - x >= 3 ? "Yes" : "No");
}
}
}
```

```
# cook your dish here
t=int(input())
for i in range(t):
x=int(input())
if x+3<=10:
print("yes")
else:
print("no")
```

```
#include <iostream>
using namespace std;
int main() {
int t;
cin>> t;
while(t--){
int x;
cin>>x;
if(10-x>=3){
cout<<"Yes";
}
else{
cout<<"No";
}
cout<<endl;
}
return 0;
}
```

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