Jiya's Sequence

 Problem

Jiya likes a sequence if all its elements when multiplied yields a number , whose least significant digit is either 2, 3 or 5.

Given the number of  test case t.The first line of each test case is a number n.Next line contains n integers - A1,A2,......An.For each given test case tell whether the given sequence will be liked by Jiya.

INPUT FORMAT-

First line constains t number of test cases.

Every test case has first line as the number of intergers the sequence contains, followed by sequence in the next line.

OUTPUT FORMAT -

Print "YES" or "NO",whether Jiya likes the sequence or not.

CONSTRAINTS-

1≤t≤100

1≤n≤15

1≤Ai≤10, for all i

 

HINT - Least significant digit of a number can be obtained by taking the number % 10.

Mind the case of output to be printed.

Sample Input

2
5
2 2 2 2 2
4
2 2 2 2

Sample Output

YES
NO

Time Limit: 1

Memory Limit: 256

Source Limit:

Explanation

For the first testcase,considering the sequence 2 2 2 2 2, when we multiply all the elements get 32 as the product.The least significant digit of 32 is 2 , so the string is liked by Jiya.

For the second testcase,considering the sequence 2 2 2 2, when we multiply all the elements get 16 as the product.The least significant digit of 16 is 6 , so the string is not liked by Jiya.

/*

// Sample code to perform I/O:

#include <iostream>

using namespace std;

int main() {

    int num;

    cin >> num;                                     // Reading input from STDIN

    cout << "Input number is " << num << endl;      // Writing output to STDOUT

}

// Warning: Printing unwanted or ill-formatted data to output will cause the test cases to fail

*/

// Write your code here

#include <iostream>

using namespace std;

int main(){

    int t;

    cin>>t;

    while(t--){

        int n;

        cin>>n;

        int arr[n];

        for(int i=0;i<n;i++){

            cin>>arr[i];

        }

        long long int mul=1;

        for(int i=0;i<n;i++){

            mul=mul*arr[i]*1LL;

        }

        int num=mul%10;

        if((num==2)||(num==3)||(num==5)){

            cout<<"YES"<<endl;

        }

        else cout<<"NO"<<endl;

    }

}