#include <iostream>
#include <vector>
#include <set>
#define repeat(i,n) for (int i = 0; (i) < int(n); ++(i))
using namespace std;
bool solve(int h, int w, vector<string> const & f) {
    auto is_on_field = [&](int y, int x) { return 0 <= y and y < h and 0 <= x and x < w; };
    auto pred = [&](int y, int x) { return is_on_field(y, x) and f[y][x] == '.'; };
    const int dy[] = { 0, -1, 0, 1 }; // counter-clockwise
    const int dx[] = { 1, 0, -1, 0 };
    int total = 0; repeat (y,h) repeat (x,w) total += pred(y, x);
    repeat (y0,h) repeat (x0,w) if (pred(y0, x0)) {
        repeat (d0,4) {
            int y = y0;
            int x = x0;
            set<pair<int, int> > used;
            for (int i = 0; i < 6; ) {
                if (not used.empty() and y == y0 and x == x0) {
                    if (used.size() == total) return true;
                    break;
                }
                int di = 0;
                for (; di < 2; ++ di) {
                    int ny = y + dy[(d0 + i + di) % 4];
                    int nx = x + dx[(d0 + i + di) % 4];
                    if (pred(ny, nx) and not used.count(make_pair(ny, nx))) {
                        y = ny;
                        x = nx;
                        i += di;
                        used.insert(make_pair(y, x));
                        break;
                    }
                }
                if (di == 2) break;
            }
        }
    }
    return false;
}
int main() {
    int h, w; cin >> h >> w;
    vector<string> f(h); repeat (y,h) cin >> f[y];
    cout << (solve(h, w, f) ? "YES" : "NO") << endl;
    return 0;
}