Problem:
Score : 100 points

Problem Statement

There is a grid with H horizontal rows and W vertical columns.
Let (i, j) denote the square at the i-th row from the top and the j-th column from the left.

For each i and j (1 \leq i \leq H, 1 \leq j \leq W), Square (i, j) is described by a character a_{i, j}.
If a_{i, j} is ., Square (i, j) is an empty square; if a_{i, j} is #, Square (i, j) is a wall square.
It is guaranteed that Squares (1, 1) and (H, W) are empty squares.

Taro will start from Square (1, 1) and reach (H, W) by repeatedly moving right or down to an adjacent empty square.

Find the number of Taro's paths from Square (1, 1) to (H, W).
As the answer can be extremely large, find the count modulo 10^9 + 7.

Constraints

H and W are integers.

2 \leq H, W \leq 1000

a_{i, j} is . or #.

Squares (1, 1) and (H, W) are empty squares.

Input

Input is given from Standard Input in the following format:

H W
a_{1, 1}\ldotsa_{1, W}
:
a_{H, 1}\ldotsa_{H, W}

Output

Print the number of Taro's paths from Square (1, 1) to (H, W), modulo 10^9 + 7.

Sample Input 1

3 4
...#
.#..
....

Sample Output 1

3

There are three paths as follows:

Sample Input 2

5 2
..
#.
..
.#
..

Sample Output 2

0

There may be no paths.

Sample Input 3

5 5
..#..
.....
#...#
.....
..#..

Sample Output 3

24

Sample Input 4

20 20
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................

Sample Output 4

345263555

Be sure to print the count modulo 10^9 + 7.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define rng(a) a.begin(),a.end()
#define rrng(a) a.rbegin(),a.rend()
#define uni(x) x.erase(unique(rng(x)),x.end())
#define PQ(T) priority_queue<T,v(T),greater<T> >
#define dup(x,y) (((x)+(y)-1)/(y))
#define v(T) vector<T>
#define vv(T) v(v(T))
#define diup(v,a) distance((v).begin(),upper_bound((v).begin(),(v).end(),(a)))
#define dilo(v,a) distance((v).begin(),lower_bound((v).begin(),(v).end(),(a)))
using ll = int64_t;
using ull = uint64_t;
using uint = uint32_t;
using pii = pair<int,int>;
using pll = pair<ll,ll>;
using pil = pair<int,ll>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
template<typename T>inline istream& operator>>(istream&i,v(T)&v)
{rep(j,v.size())i>>v[j];return i;}
template<typename T>string join(const v(T)&v)
{stringstream s;rep(i,v.size())s<<' '<<v[i];return s.str().substr(1);}
template<typename T>inline ostream& operator<<(ostream&o,const v(T)&v)
{if(v.size())o<<join(v);return o;}
template<typename T1,typename T2>inline istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<typename T1,typename T2>inline ostream& operator<<(ostream&o,const pair<T1,T2>&v)
{return o<<v.fi<<","<<v.se;}
const double eps = 1e-6;
const ll LINF = 100200300400500600ll;
const int INF = 1001001001;
//--------------------------------------------------------
#define fastio cin.tie(nullptr); std::ios::sync_with_stdio(false);
#define myio ifstream iofi; if(argc>1){iofi.open(argv[1]); std::cin.rdbuf(iofi.rdbuf());}
#define myloopset int loop = 1; if(argc>2) loop =stoi(argv[2]); rep(i,loop)
//--------------------------------------------------------------
template <class T> T chmin(T &a,T b){if(a>b){a=b; return 1;}else return 0;}
template <class T> T chmax(T &a,T b){if(a<b){a=b; return 1;}else return 0;}
const int dx[4] = {0,1,0,-1};
const int dy[4] = {-1,0,1,0};
#define MOD 1000000007

struct D21{
  int is,js;
  D21(int isize,int jsize){
    is = isize;
    js = jsize;
  }
  int operator()(int i,int j){
    return i*js+j;
  }
};

void Main(){
  int h,w;
  cin>>h>>w;
  v(string) s(h);
  cin>>s;

  vvi dp(h,vi(w,0));
  bool f = false;
  rep(i,h){
    if(s[i][0] == '#') f = true;
    if(f) dp[i][0] = 0;
    else  dp[i][0] = 1;
  }
  f = false;
  rep(j,w){
    if(s[0][j] == '#') f = true;
    if(f) dp[0][j] = 0;
    else  dp[0][j] = 1;
  }

  rep(i,h-1){
    rep(j,w-1){
      dp[i+1][j+1] = dp[i][j+1] + dp[i+1][j];
      if(dp[i+1][j+1]>MOD) dp[i+1][j+1]-=MOD;
      if(s[i+1][j+1] =='#') dp[i+1][j+1] = 0;
    }
  }
  cout<<dp[h-1][w-1]<<endl;
  
} signed main(int argc,char** argv)
{fastio; myio; myloopset Main();}