Problem:
Score : 200 points

Problem Statement

Snuke lives on an infinite two-dimensional plane. He is going on an N-day trip.
At the beginning of Day 1, he is at home. His plan is described in a string S of length N.
On Day i(1 ≦ i ≦ N), he will travel a positive distance in the following direction:

North if the i-th letter of S is N

West if the i-th letter of S is W

South if the i-th letter of S is S

East if the i-th letter of S is E

He has not decided each day's travel distance. Determine whether it is possible to set each day's travel distance so that he will be back at home at the end of Day N.

Constraints

1 ≦ | S | ≦ 1000

S consists of the letters N, W, S, E.

Input

The input is given from Standard Input in the following format:

S

Output

Print Yes if it is possible to set each day's travel distance so that he will be back at home at the end of Day N. Otherwise, print No.

Sample Input 1

SENW

Sample Output 1

Yes

If Snuke travels a distance of 1 on each day, he will be back at home at the end of day 4.

Sample Input 2

NSNNSNSN

Sample Output 2

Yes

Sample Input 3

NNEW

Sample Output 3

No

Sample Input 4

W

Sample Output 4

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"

#include <unordered_set>
#include <unordered_map>

//#include <iostream>//POJ
//#include <vector>
//#include <string>
//#include <iomanip>
//#include <math.h>
//#include <algorithm>
//#include <cstring>
using namespace std;
#define setc cin.tie(0);ios::sync_with_stdio(0)
#define dd(n) cout<<fixed<<setprecision(n)
#define inp(x) (cin>>x,x)
#define ind(x) (cin>>x,--x)
//#define repi(i,a,b) for(int i=(a), i##_len=(b); i<i##_len; ++i)
#define repi(i,a,b) for(int i=(a);i<b;++i)
#define rep(i,n) repi(i,0,n)
#define repi_(i,a,b) for(int i=(a), i##_len=(b); i<=i##_len; ++i)
#define rep_(i,n) repi_(i,0,n)
#define repir(i,a,b) for(int i=(a)-1, i##_first=(b); i>=i##_first; --i)
#define repr(i,n) repir(i,n,0)
#define repir_(i,a,b) for(int i=(a), i##_first=(b); i>=i##_first; --i)
#define repr_(i,n) repir_(i,n,0)
#define prep(i,n) for(int i=0;i*i<n;++i)
#define prepi(i,a,b) for(int i=a;i*i<b;++i)
#define all(x) (x).begin(),(x).end()
#define vsort(v) sort((v).begin(), (v).end())
#define vcsort(v,c) sort((v).begin(),(v).end(),c)
#define vrev(v) reverse((v).begin(), (v).end())
#define fnd(v,x) ((v).find(x)!=string::npos)
#define mod(a,b) (a<0?a%b+abs(b):a%b)
#define cl(a,b) (a+b-1/b)

//template<typename T> T vpop(vector<T> *v) {
//	T d = v->back();
//	v->pop_back();
//	return d;
//}
//char vpop(string *v) {
//	char d = v->back();
//	v->pop_back();
//	return d;
//}
#define siz(v) ((int)(v).size())
#define ers(v, n) (v).erase((v).begin() + n)
#define cnt(v, n) count(all(v), n)
#define vmin(v) *min_element(v.begin(), v.end())
#define vmax(v) *max_element(v.begin(), v.end())
#define contain(q) !q.empty()
#define cont(q) !q.empty()
//#define qpop(q, a, b) a=q.back().first;b=q.back().second;q.pop()
//#define pqpop(p, a, b) a=q.top().first;b=q.top().second;q.pop()
#define el "\n"
#define sp " "
#define pi 3.14159265358979
#define co(x) cout<<x<<el
#define coc(c, a, b) co(((c)?a:b))
#define cim(x) {cin>>x;--x;}
#define cim2(a,b) {cin>>a>>b;--a;--b;}
#define cosp(x) cout<<(x)<<' '
#define YES(c) coc(c,"YES", "NO")
#define Yes(c) coc(c,"Yes", "No")
#define yes(c) coc(c,"yes", "no")
#define POSSIBLE(c) coc(c, "POSSIBLE", "IMPOSSIBLE")
#define Possible(c) coc(c, "Possible", "Impossible")
#define possible(c) coc(c, "possible", "impossible")
#define inf INT_MAX
#define linf INT64_MAX*0.99
#define wildcard(T) numeric_limits<T>::min()
#define noway() {co(-1);gc();return 0;}

#define pb push_back
#define pq priority_queue
#define np next_permutation

#ifdef tqktmp_2
#define gc() getchar(); getchar()
#else
#define gc() 1
#endif
const unsigned int bf0 = (1 << 0);
const unsigned int bf1 = (1 << 1);
const unsigned int bf2 = (1 << 2);
const unsigned int bf3 = (1 << 3);
const unsigned int bf4 = (1 << 4);
const unsigned int bf5 = (1 << 5);
const unsigned int bf6 = (1 << 6);
const unsigned int bf7 = (1 << 7);
const int _10j9p7 = 1000000007;

//#define lint long long
typedef long long lint;
typedef vector<int> IV; typedef vector<string> SV;
typedef vector<lint> LIV;
typedef vector<vector<int>> IVV;
typedef pair<int,int> P; typedef pair<lint,lint> LP;
typedef vector<P> PV; typedef vector<LP> LPV;

const int
dx8[8] ={0,1,1,1,0,-1,-1,-1},
dy8[8] ={1,1,0,-1,-1,-1,0,1},
dx9[9] ={0,1,1,1,0,0,-1,-1,-1},
dy9[9] ={1,1,0,-1,0,-1,-1,0,1},
dx5[5] ={0,1,0,0,-1},
dy5[5] ={1,0,0,-1,0},
dx4[4] ={0,1,0,-1},
dy4[4] ={1,0,-1,0};
//template<class... A> void cim_(A... args) {
//	for (A *i : initializer_list<A*>{ args... }) {
//		cin>>*i;--*i;
//	}
//	return;
//}

template<class T>inline bool maxi(T &a,const T &b){ if(a < b){ a = b; return 1; } return 0; }
template<class T>inline bool mini(T &a,const T &b){ if(b < a){ a = b; return 1; } return 0; }


#define MAX_N 131082
const int md = 1e9+7;

int main(){
	setc;

	string s;cin>>s;
	bool u[4];
	lint n=siz(s);
	rep(i,n){
		if(s[i]=='N')u[0]=1;
		if(s[i]=='S')u[1]=1;
		if(s[i]=='E')u[2]=1;
		if(s[i]=='W')u[3]=1;
	}
	lint r=0;
	rep(i,4)r+=u[i];
	Yes(r%2==0);
	gc();
}