description: You had n positive integers a_1, a_2, ..., a_n arranged in a circle. For each pair of neighboring numbers (a_1 and a_2, a_2 and a_3, ..., a_{n - 1} and a_n, and a_n and a_1), you wrote down: are the numbers in the pair equal or not.
  
  Unfortunately, you've lost a piece of paper with the array a. Moreover, you are afraid that even information about equality of neighboring elements may be inconsistent. So, you are wondering: is there any array a which is consistent with information you have about equality or non-equality of corresponding pairs?
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t cases follow.
  
  The first and only line of each test case contains a non-empty string s consisting of characters E and/or N. The length of s is equal to the size of array n and 2 ≤ n ≤ 50. For each i from 1 to n: 
  
    * if s_i = E then a_i is equal to a_{i + 1} (a_n = a_1 for i = n); 
    * if s_i = N then a_i is not equal to a_{i + 1} (a_n ≠ a_1 for i = n). 
  
  Output
  
  For each test case, print YES if it's possible to choose array a that are consistent with information from s you know. Otherwise, print NO.
  
  It can be proved, that if there exists some array a, then there exists an array a of positive integers with values less or equal to 10^9.
  
  Example
  
  Input
  
  
  4
  EEE
  EN
  ENNEENE
  NENN
  
  
  Output
  
  
  YES
  NO
  YES
  YES
  
  Note
  
  In the first test case, you can choose, for example, a_1 = a_2 = a_3 = 5.
  
  In the second test case, there is no array a, since, according to s_1, a_1 is equal to a_2, but, according to s_2, a_2 is not equal to a_1.
  
  In the third test case, you can, for example, choose array a = [20, 20, 4, 50, 50, 50, 20].
  
  In the fourth test case, you can, for example, choose a = [1, 3, 3, 7].
solution: import java.io.*;
  import java.util.*;
  
  public class A {
  	public static void main(String args[]){
  		FScanner in = new FScanner();
  		PrintWriter out = new PrintWriter(System.out);
  		int t = in.nextInt();
  	m:	while(t-->0) { 
  			char arr[]=in.next().toCharArray();
  			int c=0,pos=-1;
  
  			if(arr.length==2)
  			{
  				if(arr[0]!=arr[arr.length-1])
  					out.println("NO");
  				else
  					out.println("YES");
  				continue m;
  			}
  
  			for(int i=0;i<arr.length;i++)
  			{
  				char x=arr[i];
  				if(x=='N')
  				{
  					c++;
  					pos=i;
  				}
  			}
  
  			if(c==1)
  			{
  				out.println("NO");
  			}
  			else
  				out.println("YES");
  		}
  		out.close();
  	}
  
  	static void sort(int[] a) {
  		ArrayList<Integer> l=new ArrayList<>();
  		for (int i:a) l.add(i);
  		Collections.sort(l);
  		for (int i=0; i<a.length; i++) a[i]=l.get(i);
  	}
  
  	static boolean checkprime(int n1)
  	{
  		if(n1%2==0||n1%3==0)
  			return false;
  		else 
  		{
  			for(int i=5;i*i<=n1;i+=6)
  			{
  				if(n1%i==0||n1%(i+2)==0)
  					return false;
  			}
  			return true;
  		}
  	}
  	
  	static class FScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer sb = new StringTokenizer("");
  
  		String next(){
  			while(!sb.hasMoreTokens()){
  				try{
  					sb = new StringTokenizer(br.readLine());
  				} catch(IOException e){ }
  			}
  			return sb.nextToken();
  		}
  		String nextLine(){
  			try{ return br.readLine(); } 
  			catch(IOException e) { } return "";
  		}
  
  		int nextInt(){
  			return Integer.parseInt(next());
  		}
  
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		int[] readArray(int n) {
  			int a[] = new int[n];
  			for(int i=0;i<n;i++) a[i] = nextInt();
  			return a;
  		}
  
  		float nextFloat(){
  		return Float.parseFloat(next());
  		}
  		
  		double nextDouble(){
  		return Double.parseDouble(next());
  		}
  	}
  }
  