A. Equal or Not Equal

A. Equal or Not Equal

time limit per test:2 seconds

memory limit per test:256 megabytes

input:standard input

output:standard output

You had nn positive integers a1,a2,…,ana1,a2,…,an arranged in a circle. For each pair of neighboring numbers (a1a1 and a2a2, a2a2 and a3a3, ..., an−1an−1 and anan, and anan and a1a1), you wrote down: are the numbers in the pair equal or not.

Unfortunately, you've lost a piece of paper with the array aa. Moreover, you are afraid that even information about equality of neighboring elements may be inconsistent. So, you are wondering: is there any array aa which is consistent with information you have about equality or non-equality of corresponding pairs?

Input

The first line contains a single integer tt (1≤t≤10001≤t≤1000) — the number of test cases. Next tt cases follow.

The first and only line of each test case contains a non-empty string ss consisting of characters E and/or N. The length of ss is equal to the size of array nn and 2≤n≤502≤n≤50. For each ii from 11 to nn:

• if si=si= E then aiai is equal to ai+1ai+1 (an=a1an=a1 for i=ni=n);
• if si=si= N then aiai is not equal to ai+1ai+1 (an≠a1an≠a1 for i=ni=n).

Output

For each test case, print YES if it's possible to choose array aa that are consistent with information from ss you know. Otherwise, print NO.

It can be proved, that if there exists some array aa, then there exists an array aa of positive integers with values less or equal to 109109.

Example

input

4
EEE
EN
ENNEENE
NENN

output

YES
NO
YES
YES

Note

In the first test case, you can choose, for example, a1=a2=a3=5a1=a2=a3=5.

In the second test case, there is no array aa, since, according to s1s1, a1a1 is equal to a2a2, but, according to s2s2, a2a2 is not equal to a1a1.

In the third test case, you can, for example, choose array a=[20,20,4,50,50,50,20]a=[20,20,4,50,50,50,20].

In the fourth test case, you can, for example, choose a=[1,3,3,7]a=[1,3,3,7].

思路：题目说有每一个例子都是首尾相连的。在每一个例子中，下一个与该个相等为E，不相等为N，也就是说，如果首位成环之后是否有存在一种序列使得这个环成立。该题思路是：因为能使得数变化的只有N，既然要成环的话，自然就说明在这个例子中，N不能只有一个，N的数量如果>=2的话，怎么都是可以找到能成立的序列的。所以这题只需要判断给出的例子是否只有一个N，如果只有一个就是NO，如果有多个就是YES。

题解：

#include<bits/stdc++.h>
using namespace std;
int n;
string s;
int main(){
	cin>>n;
	while(n--){
		cin>>s;
		int a=0;
		for(int i=0;i<s.length();i++){
			if(s[i]=='N'){
				a++;
			}
		}if(a==1){
			cout<<"NO"<<endl;
		}else{
			cout<<"YES"<<endl;
		}
	}
	return 0;
}