#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <chrono>
#include <climits>
#include <cmath>
#include <complex>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <vector>
using namespace std;

#define ll long long
#define forn(i, s, n) for (int i = s; i < n; i++)
#define bforn(i, s) for (int i = s; i >= 0; i--)
// for pairs
#define s second
#define f first
const int M = 2e5 + 1;
const char nl = '\n';
const int E = 1505;

int g[E][E];
bool blocked;

void ff(int i, int j){
    if(i < 0 || i >= E || j < 0 || j >= E) return;
    if (g[i][j] == 1 || g[i][j] == 2) return;
    if (g[i][j] == -1){
        blocked = false;
        return;
    }
    g[i][j] = 2;
    ff(i + 1, j);
    ff(i - 1, j);
    ff(i, j - 1);
    ff(i, j + 1);
}

void solve(){
    int n; cin >> n;
    string s; cin >> s;
    for (int i = 0; i < E; i++){
        for (int j = 0; j < E; j++){
            if (i == 0 || i == E || j == 0 || j == E) g[i][j] = -1;
            else g[i][j] = 0;
        }
    }
    int i = 1000, j = 1000;
    g[i][j] = 1;
    for (int k = 0; k < s.size(); k++){
        g[i][j] = 1;
        if (s[k] == 'N'){
            g[i + 1][j] = 1;
            g[i + 2][j] = 1;
            i+=2;
        }
        if (s[k] == 'W'){
            g[i][j - 1] = 1;
            g[i][j - 2] = 1;
            j -= 2;
        }
        if (s[k] == 'E'){
            g[i][j + 1] = 1;
            g[i][j + 2] = 1;
            j+=2;
        }
        if (s[k] == 'S'){
            g[i - 1][j] = 1;
            g[i - 2][j] = 1;
            i -= 2;
        }
    }
    int ans = 0;
    for (int k = 0; k < E; k++){
        for (int l = 0; l < E; l++){
            if (g[k][l] != 0) continue;
            blocked = true;
            ff(k, l);
            if (blocked) ans++;
        }
    }
    cout << ans << nl;
}
// if you are unsure of how to do a problem by hand
// the solution is most likely somewhat brute force (i.e binsearch)
// if input is limited to 0-9 or a-z, usually you will loop through all combinations
// of letters or numbers
// constructive problems : start with 1 when constructing
// :%y+ to copy all lines
// rearrange math expressions
// difference array? (a[i] - b[i])
// debugging : reread the problem statement-every word is important!
// debugging : print important variables!
// think brute force sol first
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    freopen("gates.in", "r", stdin);
    freopen("gates.out", "w", stdout);
    solve();
}