Chef got into a fight with the evil Dr Doof. Dr Doof has decided to destroy all even numbers from the universe using his Evil-Destroy-inator. Chef has $N$ integers with him. To stop Doof, Chef has to find an odd number which is an integer multiple of all $N$ numbers that he has with him. Find if it is possible for Chef to prevent Dr Doof from destroying the even numbers.
Formally, given $N$ positive integers, find if there exists an odd number which is an integer multiple of all the given $N$ numbers. If yes, print "YES", otherwise "NO". You can print any letter in any case.
-----Input-----
- First line contains $T$, number of testcases. Each testcase consists of $2$ lines.
- The first line of each test case consists of a positive integer $N$, denoting the number of positive integers Chef has.
- The second line of each test case contains $N$ space separated integers $A_i$ each denoting an integer that Chef has with him.
-----Output-----
For every test case, if there exists such an odd number, print "YES" on a separate line, otherwise "NO". The judge is case insensitive. That means, your code can print any letter in any case ( "Yes", "yes" or "YES" are all accepted).
-----Constraints-----
- $1 \leq T \leq 10^3$
- $1 \leq N \leq 10^3$
- $1 \leq A_i \leq 10^3$
-----Sample Input-----
2
5
1 2 5 4 3
1
7
-----Sample Output-----
NO
YES
-----Explanation-----
For test $1$: There exists no odd number.
For test $2$: The possible odd numbers can be $7$, $21$, $49$, $315$, …
# cook your dish here
for t in range(int(input())):
 n=int(input())
 arr=list(map(int,input().split()[:n]))
 c=1
 for i in arr:
  c=c*i
 if c%2!=0:
  print("YES")
 else:
  print("NO")
# cook your dish here
for t in range(int(input())):
 n=int(input())
 arr=list(map(int,input().split()[:n]))
 c=1
 for i in arr:
  c=c*i
 if c%2!=0:
  print("YES")
 else:
  print("NO")