#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; typedef pair pli; typedef pair pil; typedef pair pll; typedef pair pii; typedef pair pdd; typedef pair piii; typedef pair pil; typedef pair plii; typedef pair pdi; typedef long long ll; typedef unsigned long long ull; typedef pair puu; typedef long double ld; const int N = 2000086, MOD = 998244353, INF = 0x3f3f3f3f, MID = 333; const long double EPS = 1e-8; int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1}; // int dx[8] = {1, 1, 0, -1, -1, -1, 0, 1}, dy[8] = {0, 1, 1, 1, 0, -1, -1, -1}; // int dx[8] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[8] = {1, 2, 2, 1, -1, -2, -2, -1}; int n, m, cnt; int w[N]; vector num; ll res; ll lowbit(ll x) { return x & -x; } ll lcm(ll a, ll b) { return a / __gcd(a, b) * b; } inline double rand(double l, double r) { return (double)rand() / RAND_MAX * (r - l) + l; } inline ll qmi(ll a, ll b, ll c) { ll res = 1; while (b) { if (b & 1) res = res * a % c; a = a * a % c; b >>= 1; } return res; } inline ll qmi(ll a, ll b) { ll res = 1; while (b) { if (b & 1) res *= a; a *= a; b >>= 1; } return res; } inline double qmi(double a, ll b) { double res = 1; while (b) { if (b & 1) res *= a; a *= a; b >>= 1; } return res; } inline ll C(ll a, ll b, int* c) { if (a < b) return 0; ll res = 1; for (ll j = a, i = 1; i < b + 1; i++, j--) res *= j; for (ll j = a, i = 1; i < b + 1; i++, j--) res /= i; return res; } inline int find_(int x) { return lower_bound(num.begin(), num.end(), x) - num.begin(); } char s[408][408]; ll f[2][408][408]; void solve() { f[1][1][2] = 1; for (int i = 3; i < n + m - 1; i++) { memset(f[i & 1 ^ 1], 0, sizeof f[0]); for (int j = 1; j <= n; j++) for (int k = 1; k <= n; k++) if (f[i & 1][j][k] && s[j][i - j] != '#' && s[k][i - k] != '#') { int x1 = j, x2 = k, y1 = i - j, y2 = i - k; if (x1 != n) { if (x2 != n) f[i & 1 ^ 1][x1 + 1][x2 + 1] = (f[i & 1 ^ 1][x1 + 1][x2 + 1] + f[i & 1][j][k]) % MOD; if (y2 != m && x1 + 1 != x2) f[i & 1 ^ 1][x1 + 1][x2] = (f[i & 1 ^ 1][x1 + 1][x2] + f[i & 1][j][k]) % MOD; } if (y1 != m) { if (x2 != n && x1 != x2 + 1) f[i & 1 ^ 1][x1][x2 + 1] = (f[i & 1 ^ 1][x1][x2 + 1] + f[i & 1][j][k]) % MOD; if (y2 != m) f[i & 1 ^ 1][x1][x2] = (f[i & 1 ^ 1][x1][x2] + f[i & 1][j][k]) % MOD; } } } printf("%lld\n", (f[(n + m - 1) & 1][n][n - 1] + f[(n + m - 1) & 1][n - 1][n]) % MOD); } int main() { cin >> n >> m; for (int i = 1; i < n + 1; i++) scanf("%s", s[i] + 1); if (s[1][1] == '#' || s[1][2] == '#' || s[2][1] == '#' || s[n][m] == '#' || s[n - 1][m] == '#' || s[n][m - 1] == '#') { puts("0"); return 0; } solve(); return 0; }