#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    string seq;
    cin >> seq;
    int count = 0;
    for (int i = 0; i < seq.length(); i++)
      if (seq[i] == 'N') count++;
    if (count == 1)
      cout << "NO"
           << "\n";
    else
      cout << "YES"
           << "\n";
  }
}

### Prompt

Create a solution in Cpp for the following problem:

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].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    string seq;
    cin >> seq;
    int count = 0;
    for (int i = 0; i < seq.length(); i++)
      if (seq[i] == 'N') count++;
    if (count == 1)
      cout << "NO"
           << "\n";
    else
      cout << "YES"
           << "\n";
  }
}

```