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.BufferedReader; 
  import java.io.IOException; 
  import java.io.InputStreamReader; 
  import java.util.Scanner; 
  import java.util.StringTokenizer;
  import java.util.*;
  public class codeforcesA{
     static class FastReader 
      { 
          BufferedReader br; 
          StringTokenizer st; 
   
          public FastReader() 
          { 
              br = new BufferedReader(new
                       InputStreamReader(System.in)); 
          } 
   
          String next() 
          { 
              while (st == null || !st.hasMoreElements()) 
              { 
                  try
                  { 
                      st = new StringTokenizer(br.readLine()); 
                  } 
                  catch (IOException  e) 
                  { 
                      e.printStackTrace(); 
                  } 
              } 
              return st.nextToken(); 
          } 
   
          int nextInt() 
          { 
              return Integer.parseInt(next()); 
          } 
   
          long nextLong() 
          { 
              return Long.parseLong(next()); 
          } 
   
          double nextDouble() 
          { 
              return Double.parseDouble(next()); 
          } 
   
          String nextLine() 
          { 
              String str = ""; 
              try
              { 
                  str = br.readLine(); 
              } 
              catch (IOException e) 
              { 
                  e.printStackTrace(); 
              } 
              return str; 
           
     } 
  }
  static int gcd(int a,int b){if(b==0){return a;}return gcd(b,a%b);}
  
    public static void main(String args[]){
       FastReader sc=new FastReader();
       int t=sc.nextInt();
       while(t-->0){
       StringBuilder sb=new StringBuilder();
       String s=sc.nextLine();
        int n=s.length();
        int e=0,ne=0;
        for(int i=0;i<n;i++){if(s.charAt(i)=='E'){e++;}else{ne++;}}
          if( (ne==1  )){System.out.println("NO");}
          else{System.out.println("YES");}
          }
       }
   }
   
  