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.lang.reflect.Array;
  import java.util.Arrays;
  import java.util.Scanner;
  
  public class EqualOrNotEqual {
      public static void main(String[] args) {
          Scanner scanner = new Scanner(System.in);
          int t = scanner.nextInt();
          while (t-->0){
              String s = scanner.next();
              int[] a = new int[s.length()];
              int num = 1;
              int index = s.length()-1;
              int slen = s.length();
              if(s.charAt(slen-1) == 'E'){
                  a[index--] = num;
                  a[0] = num;
              }else{
                  a[index--] = num;
                  a[0] = ++num;
              }
              boolean FoundN = false;
              int lastNIndex = 0;
              int nCount = 0;
              for (int i = s.length()-2; i >=0; i--) {
                  if (s.charAt(i) == 'E'){
                      a[index--]= a[index+2];
                  }else {
                      if(!FoundN){
                          FoundN = true;
                          lastNIndex = i;
                      }
                      nCount++;
                      a[index--] = ++num;
                  }
              }
  
              if(FoundN && s.charAt(s.length()-1) != 'N' && nCount > 1){
                  int i = s.length()-1;
                  while (i>lastNIndex){
                      a[i] = a[0];
                      i--;
                  }
              }
  
  
  
  //            if(s.charAt(s.length()-2) == 'N' && s.charAt(s.length()-1) == 'E'){
  //                a[a.length-1] = a[0];
  //            }
  
  
  //            System.out.println(Arrays.toString(a));
  //            if ((s.charAt(0) == 'E' && a[0] != a[1]) || (s.charAt(0) == 'N' && a[0] == a[1]) ){
              if((s.charAt(s.length()-1) == 'E' && a[0] != a[a.length-1]) || (s.charAt(s.length()-1) == 'N' && a[0] == a[a.length-1])){
                  System.out.println("NO");
              }else {
                  System.out.println("YES");
              }
  
  //            if ('N' == s.charAt(s.length()-2)){
  //                System.out.println("YES");
  //            }else {
  //                System.out.println("NO");
  //            }
          }
      }
  }
  