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: /*
          "Everything in the universe is balanced. Every disappointment
                  you face in life will be balanced by something good for you!
                          Keep going, never give up."
  
  */
  
  
  
  import java.util.*;
  import java.lang.*;
  import java.io.*;
  
  public class Solution {
      public static void main(String[] args) throws java.lang.Exception {
          out = new PrintWriter(new BufferedOutputStream(System.out));
          sc = new FastReader();
  
          int test = sc.nextInt();
          for (int t = 0; t < test; t++) {
              solve();
          }
          out.close();
      }
  
      private static void solve() {
          char[] arr = sc.next().toCharArray();
          int n = arr.length;
          int count = 0;
          for (char c : arr) {
              if (c == 'N') {
                  count++;
              }
          }
          if (count == 1) {
              out.println("NO");
          }else {
              out.println("YES");
          }
      }
  
  
      public static FastReader sc;
      public static PrintWriter out;
      static class FastReader
      {
          BufferedReader br;
          StringTokenizer str;
  
          public FastReader()
          {
              br = new BufferedReader(new
                      InputStreamReader(System.in));
          }
  
          String next()
          {
              while (str == null || !str.hasMoreElements())
              {
                  try
                  {
                      str = new StringTokenizer(br.readLine());
                  }
                  catch (IOException  end)
                  {
                      end.printStackTrace();
                  }
              }
              return str.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 end)
              {
                  end.printStackTrace();
              }
              return str;
          }
      }
  }