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.StringTokenizer;
  
  public class EqualOrNotEqual {
      public static void main(String[] args) {
          
          FastReader fr = new FastReader();
          
          // ENE
          // a1 equal to a2, a2 not equal to a3, a3 can be equal to a1
          // the inconsistency comes when the loop completes, what is the condition?
  
          // if last is N
          // then there can be no line of Es from 1 to n. There must be a not equal somewhere
  
          // if last is E
          // there can not be a line of Ns from 1 to n. There must be 
  
  
          // a1 == a2, a2 != a3, therefore a3 != a1 which is a contradition to the proposed
  
          // EEEEEE = fine
          // EEENEE = not fine
          // EEENNE = fine
          // EENNNE = fine
  
          // there must be either 0, 2 or more Ns, then we are fine
  
          int ts = fr.nextInt();
          StringBuilder result = new StringBuilder();
  
          for (int t = 0; t < ts; t++) {
  
              int NCount = 0;
  
              String s = fr.nextLine();
  
              for (char c : s.toCharArray()) {
                  if (c == 'N') {
                      NCount++;
                      if (NCount >= 2) {
                          break;
                      }
                  }
              }
  
              result.append(NCount == 1 ? "NO" : "YES");
              result.append("\n");
  
          }
  
          System.out.println(result.toString());
  
  
      }
  
  
      /** HELPER */
  
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
   
          public FastReader()
          {
              br = new BufferedReader(
                  new InputStreamReader(System.in));
          }
  
          int[] nextLineAsIntArray(int n) {
              int[] result = new int[n];
  
              for (int i = 0; i < n; i++) {
                  result[i] = nextInt();
              }
              return result;
          }
  
          long[] nextLineAsLongArray(int n) {
              long[] result = new long[n];
  
              for (int i = 0; i < n; i++) {
                  result[i] = nextLong();
              }
  
              return result;
          }
   
          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;
          }
      }
  }
  