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.util.*;
  import java.lang.*;
  import java.io.*;
  public class Solution{
      public static void main (String[] args) throws java.lang.Exception {
          int t = sc.nextInt();
          while(t-->0){
              String str = sc.nextLine();
              int c=0;
              for(int i=0; i<str.length(); i++){
                  if(str.charAt(i)=='E') c++;
              }
              c++;
              if(c==str.length()) System.out.println("NO");
              else System.out.println("YES");
  
          }
      }
      // Fast Reader Class
      public static FastReader sc = new FastReader();
      public static BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
      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; }
      }
  }
  //    int n = sc.nextInt(), p = sc.nextInt(), k = sc.nextInt();
  //            if(p%k<=(n-1)%k){
  //                    n--;
  //                    long ans = (p%k)*((n+k)/k)-((p%k-1)*(p%k))/(2*k);
  //                    System.out.println(ans);
  //                    ans+=(p-(p%k))/k+1;
  //                    System.out.println(ans);
  //                    }else{
  //                    long ans=0;
  //                    n--;
  //                    ans+=(n%k)*(n+k)/k;
  //                    ans+=(p%k-n%k-1)*(((n+k)/k)-1);
  //                    ans+=(p-(p%k))/k+1;
  //                    System.out.println(ans);
  //                    }