func_code_string: #include <bits/stdc++.h>   using namespace std;   template <class T, class U>   ostream& operator<<(ostream& os, const pair<T, U>& p) {     os <<  (  << p.first <<  ,   << p.second <<  ) ;     return os;   }   template <class T>   ostream& operator<<(ostream& os, const vector<T>& v) {     os <<  { ;     for (int i = 0; i < (int)v.size(); i++) {       if (i) os <<  ,  ;       os << v[i];     }     os <<  } ;     return os;   }   template <typename Arg1>   void __f(const char* name, Arg1&& arg1) {     cerr << name <<   :   << arg1 << std::endl;   }   template <typename Arg1, typename... Args>   void __f(const char* names, Arg1&& arg1, Args&&... args) {     const char* comma = strchr(names + 1,  , );     cerr.write(names, comma - names) <<   :   << arg1 <<   |  ;     __f(comma + 1, args...);   }   const int mod = 1e9 + 7;   const int N = 2005;   inline int add(int x, int y) {     x += y;     if (x >= mod) x -= mod;     return x;   }   inline int sub(int x, int y) {     x -= y;     if (x < 0) x += mod;     return x;   }   inline int mul(int x, int y) { return (x * 1ll * y) % mod; }   inline int powr(int a, long long b) {     int x = 1 % mod;     while (b) {       if (b & 1) x = mul(x, a);       a = mul(a, a);       b >>= 1;     }     return x;   }   inline int inv(int a) { return powr(a, mod - 2); }   char s[N][N];   int dp[2][N][N], dp2[2][N][N];   int num[2][N][N];   int main() {     int n = 2000, m = 2000;     scanf( %d , &(n)), scanf( %d , &(m));     for (int i = 1; i <= n; i++) {       scanf( %s , s[i] + 1);     }     for (int i = n; i >= 1; i--) {       for (int j = m; j >= 1; j--) {         num[0][i][j] = num[0][i][j + 1] + (s[i][j] ==  R );         num[1][i][j] = num[1][i + 1][j] + (s[i][j] ==  R );         if (i == n) dp[0][i][j] = (num[0][i][j] == 0);         if (j == m) dp[1][i][j] = (num[1][i][j] == 0);         int mxj = m - num[0][i][j];         int mxi = n - num[1][i][j];         if (i != n) dp[0][i][j] = sub(dp2[1][i + 1][j], dp2[1][i + 1][mxj + 1]);         if (j != m) dp[1][i][j] = sub(dp2[0][i][j + 1], dp2[0][mxi + 1][j + 1]);         dp2[0][i][j] = add(dp[0][i][j], dp2[0][i + 1][j]);         dp2[1][i][j] = add(dp[1][i][j], dp2[1][i][j + 1]);       }     }     assert(dp[0][1][1] == dp[1][1][1]);     printf( %d n , dp[0][1][1]);   }   