#include <bits/stdc++.h> using namespace std; #if __has_include(<atcoder/all>) #include <atcoder/all> using namespace atcoder; template<int mod>istream &operator>>(istream &is,static_modint<mod> &a){long long b;is>>b;a=b;return is;} istream &operator>>(istream &is,modint &a){long long b;cin>>b;a=b;return is;} #endif using ll=long long; using ull=unsigned long long; using P=pair<ll,ll>; template<typename T>using minque=priority_queue<T,vector<T>,greater<T>>; template<typename T>bool chmax(T &a,const T &b){return (a<b?(a=b,true):false);} template<typename T>bool chmin(T &a,const T &b){return (a>b?(a=b,true):false);} template<typename T1,typename T2>istream &operator>>(istream &is,pair<T1,T2>&p){is>>p.first>>p.second;return is;} template<typename T>istream &operator>>(istream &is,vector<T> &a){for(auto &i:a)is>>i;return is;} template<typename T1,typename T2>void operator++(pair<T1,T2>&a,int n){a.first++,a.second++;} template<typename T1,typename T2>void operator--(pair<T1,T2>&a,int n){a.first--,a.second--;} template<typename T>void operator++(vector<T>&a,int n){for(auto &i:a)i++;} template<typename T>void operator--(vector<T>&a,int n){for(auto &i:a)i--;} #define reps(i,a,n) for(int i=(a);i<(n);i++) #define rep(i,n) reps(i,0,n) #define all(x) x.begin(),x.end() #define pcnt(x) __builtin_popcount(x) ll myceil(ll a,ll b){return (a+b-1)/b;} #ifdef LOCAL #include "debug.h" #else #define debug(...) static_cast<void>(0) template<typename T1,typename T2>ostream &operator<<(ostream &os,const pair<T1,T2>&p){os<<p.first<<' '<<p.second;return os;} #endif void SOLVE(); int main(){ ios::sync_with_stdio(false); cin.tie(nullptr); #ifdef LOCAL clock_t start=clock(); #endif int testcase=1; //cin>>testcase; for(int i=0;i<testcase;i++){ SOLVE(); } #ifdef LOCAL cout<<"time:"; cout<<(clock()-start)/1000; cout<<"ms\n"; #endif } using mint=modint998244353; void SOLVE(){ int h,w,kk; cin>>h>>w>>kk; vector<string>s(h); cin>>s; vector dp(h,vector<vector<mint>>(w,vector<mint>(kk,0))); dp[0][0][0]=1; rep(i,h)rep(j,w)if(s[i][j]!='#'){ rep(k,kk){ if(s[i][j]=='o'){ if(i&&k)dp[i][j][k]+=dp[i-1][j][k-1]; if(j&&k)dp[i][j][k]+=dp[i][j-1][k-1]; } else{ if(i)dp[i][j][k]+=dp[i-1][j][k]; if(j)dp[i][j][k]+=dp[i][j-1][k]; } } } mint ans=0; rep(i,kk)ans+=dp[h-1][w-1][i]; cout<<ans.val()<<endl; }