#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
using namespace std;
#define ll long long
#define fff(i,a,b) for(ll i = a; i < b; i++)
ll n, m;
vector<ll> adj[501][501];
ll dir[501][501];
ll table[257];
priority_queue<pair<ll, pair<ll, ll>>> q;
set<pair<ll, ll>> seen;
pair<ll, ll> adjPos[]{{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
void setIO(string s) {
freopen((s + ".in").c_str(), "r", stdin);
freopen((s + ".out").c_str(), "w", stdout);
}
int main(){
// setIO("testing");
// opposite of input, for my own clarity
cin >> n >> m;
// NESW, so if i have W and want to know how many i need to do to get N it should be
table['N'] = 0;
table['E'] = 1;
table['S'] = 2;
table['W'] = 3;
table['X'] = -1;
fff(i, 0, n){
fff(j, 0, m){
char c; cin >> c;
dir[i][j] = table[c];
}
}
q.push({0, {0, 0}});
while (q.size()){
auto [d, pos] = q.top(); q.pop();
auto [y, x] = pos;
if (seen.find({y, x}) != seen.end()) continue;
// process this spot.
seen.insert({y, x});
if (y == n-1 && x == m-1){
// at the end, output dist
cout << -d << endl;
return 0;
}
// now, move to nearby, if can
if (dir[y][x] == -1) continue;
fff(i, 0, 4){
auto [dy, dx] = adjPos[i];
ll ny = y+dy, nx = x+dx;
if (ny < 0 || ny >= n || nx < 0 || nx >= m) continue;
auto cost = -((i - dir[y][x]+4)%4);
q.push({cost+d, {ny, nx}});
}
}
cout << -1 << endl;
}
