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)

#ifndef LOCAL
#pragma GCC optimize("O3")
#endif
#include "bits/stdc++.h"
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;
#define sim template < class c
#define ris return * this
#define dor > debug & operator <<
#define eni(x) sim > typename \
  enable_if<sizeof dud<c>(0) x 1, debug&>::type operator<<(c i) {
sim > struct rge { c b, e; };
sim > rge<c> range(c i, c j) { return rge<c>{i, j}; }
sim > auto dud(c* x) -> decltype(cerr << *x, 0);
sim > char dud(...);
struct debug {
#ifdef LOCAL
~debug() { cerr << endl; }
eni(!=) cerr << boolalpha << i; ris; }
eni(==) ris << range(begin(i), end(i)); }
sim, class b dor(pair < b, c > d) {
  ris << "(" << d.first << ", " << d.second << ")";
}
sim dor(rge<c> d) {
  *this << "[";
  for (auto it = d.b; it != d.e; ++it)
    *this << ", " + 2 * (it == d.b) << *it;
  ris << "]";
}
#else
sim dor(const c&) { ris; }
#endif
};
#define imie(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "
#define fastio ios_base::sync_with_stdio(false);cin.tie(0);
typedef long long ll;
typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> indexed_set; 
const ll MAXN = 1010; 
const ll MOD = 1e9 + 7; 

ll n , m ;
char grid[MAXN][MAXN] ; 
ll dp[MAXN][MAXN] ; 
ll dfs(ll x,ll y){
	if ( x >= n || y >= m ) return 0 ; 
	if ( x == n-1 && y == m-1 ){
		return 1 ; 
	}
	if ( grid[x][y] == '#' ) return 0 ; 
	if ( dp[x][y] != -1 ) return dp[x][y] ; 
	ll ans = 0 ;
	ans += (dfs(x+1,y) + dfs(x,y+1)) % MOD  ; 
	return dp[x][y] = ans % MOD ; 
}
int main(){
    fastio
    memset(dp,-1,sizeof dp) ; 
    cin >> n >> m ;
    for(ll i=0;i<n;i++){
    	for(ll j=0;j<m;j++){
    		cin >> grid[i][j] ;
    	}
    }
    cout << dfs(0,0) << endl;
    return 0 ;
}