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: #include <bits/stdc++.h>
  using namespace std;
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long extgcd(long long a, long long b, long long& x, long long& y) {
    if (b == 0) {
      x = 1;
      y = 0;
      return a;
    }
    long long x1, y1;
    long long d = extgcd(b, a % b, x1, y1);
    x = y1;
    y = x1 - y1 * (a / b);
    return d;
  }
  long long revrsno(long long n) {
    long long k = 0;
    vector<long long> z;
    while (n > 0) {
      z.push_back(n % 10);
      n /= 10;
    }
    for (long long i = 0; i < z.size(); i++) k = k * 10 + z[i];
    return k;
  }
  bool isPrime(long long n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    for (long long i = 5; i * i <= n; i = i + 6)
      if (n % i == 0 || n % (i + 2) == 0) return false;
    return true;
  }
  bool isPowerOfTwo(long long n) {
    if (n == 0) return false;
    return (ceil(log2(n)) == floor(log2(n)));
  }
  long long powm(long long a, long long b) {
    long long res = 1;
    while (b) {
      if (b & 1) res = (res * a) % 1000000007;
      a = (a * a) % 1000000007;
      b >>= 1;
    }
    return res;
  }
  long long divide(long long a, long long b) {
    return (a % 1000000007) * powm(b, 1000000007 - 2) % 1000000007;
  }
  long long mul(long long a, long long b) {
    return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
  }
  long long add(long long a, long long b) {
    return (a % 1000000007 + b % 1000000007) % 1000000007;
  }
  long long Totfun(long long n) {
    long long z = n;
    if (n % 2 == 0) {
      while (n % 2 == 0) {
        n /= 2;
      }
      z /= 2;
    }
    for (long long i = 3; i <= sqrt(n); i += 2) {
      if (isPrime(i) && n % i == 0) {
        while (n % i == 0) {
          n /= i;
        }
        z -= z / i;
      }
    }
    if (n > 1) {
      z -= z / n;
    }
    return z;
  }
  long long maxPrimeFactor(long long n) {
    long long maxPrime = -1;
    while (n % 2 == 0) {
      maxPrime = 2;
      n >>= 1;
    }
    for (long long i = 3; i <= sqrt(n); i += 2) {
      while (n % i == 0) {
        maxPrime = i;
        n = n / i;
      }
    }
    if (n > 2) maxPrime = n;
    return maxPrime;
  }
  long long fact[100001];
  long long nCr(long long n, long long r) {
    if (r > n || r < 0) return 0;
    long long z = fact[n];
    z = mul(z, powm(fact[n - r], 1000000007 - 2));
    z = mul(z, powm(fact[r], 1000000007 - 2));
    return z;
  }
  void solve() {
    string s;
    cin >> s;
    long long n = (int)(s).size();
    long long no = 0;
    for (long long i = 0; i < n; i++)
      if (s[i] == 'N') no++;
    if (no == 0 || no >= 2)
      cout << "YES";
    else
      cout << "NO";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t;
    cin >> t;
    while (t--) {
      solve();
      cout << endl;
    }
    return 0;
  }
  