#include
#include
using namespace std;
int main() {
int n, m;
cin >> n >> m;
char mat[n][m];
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
cin >> mat[i][j];
}
}
// prvo da najdeme kade se naoga S i kade se naoga E
int si = -1, sj = -1;
int ei = -1, ej = -1;
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
if(mat[i][j] == 'S') {
si = i;
sj = j;
}
if(mat[i][j] == 'E') {
ei = i;
ej = j;
}
}
}
bool visited[n][m]; // ke ni kazuva koi polinja sme gi posetile a koi ne
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
visited[i][j] = false;
}
}
queue q;
q.push(si);
q.push(sj);
q.push(0); // kolku cekori sum napravil od poleto S do momentalnoto pole
visited[si][sj] = true;
int di[] = {-1, 1, 0, 0};
int dj[] = {0, 0,-1, 1};
while(!q.empty()) {
int ci = q.front();
q.pop();
int cj = q.front();
q.pop();
int cekori = q.front();
q.pop();
if(mat[ci][cj] == 'E') { // ova znaci deka sme stignale do krajot odnosno do posakuvanata destinacija
cout << "Od S do E ni trebaat " << cekori << " broj na cekori" << endl;
break;
}
for(int i = 0; i < 4; i++) {
int ti = ci + di[i];
int tj = cj + dj[i];
if(ti >= 0 and ti < n and tj >= 0 and tj < m and mat[ti][tj] != '#' and !visited[ti][tj]) {
q.push(ti);
q.push(tj);
q.push(cekori + 1);
visited[ti][tj] = true;
}
}
}
return 0;
}
/*
5 5
.....
#####
S....
E....
#####
5 5
#..E#
.....
##...
S....
#####
Queue: (3, 0)
(3, 0) --> (3, 1)
Queue: (3, 1)
(3, 1) --> (3, 2)
Queue: (3, 2) --> (3, 3); (2, 2)
Queue(3, 3); (2, 2)
(3, 3) --> (3, 4); (2, 3)
Queue: (2, 2), (3, 4), (2, 3)
(2, 2) --> (1, 2)
Queue: (3, 4), (2, 3), (1, 2)
(3, 4) --> (2, 4)
Queue: (2, 3), (1, 2), (2, 4)
(2, 3) --> (1, 3)
Queue: (1, 2), (2, 4), (1, 3)
(1, 2) --> (1, 1); (0, 2)
Queue: (2, 4); (1, 3); (1, 1); (0, 2)
(2, 4) --> (1, 4)
Queue: (1, 3); (1, 1); (0, 2); (1, 4)
(1, 3) --> (0, 3)
*/