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: Runtime Error)

/*author : Yashvardhan
Wednesday, January 30, 2019
10:55 PM
*/

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

#define int ll
#define pb push_back
#define vi vector<int>
#define pi pair<int,int>
#define vpi vector<pi>
#define ff first
#define ss second
#define endl '\n'

#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize ("O3")
#pragma GCC optimize ("O2")
#pragma GCC optimize("Ofast")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define _CRT_SECURE_NO_WARNINGS

#ifdef __APPLE__
#define debug(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){cerr << name << " : " << arg1 << 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...);}
#else
#define debug(...)
#endif

template <typename T> ostream& operator << (ostream& os,const vector<T>& v){for(int i=0;i<v.size();++i) os<<v[i]<<" ";return os;} 
template <typename T> ostream& operator << (ostream& os,const set<T>& v){for(auto it:v) os<<it<<" ";return os;} 
template <typename T, typename S> ostream& operator << (ostream& os, const pair<T,S>& v) {os<<v.ff<<" "<<v.ss;return os;}

const int mod = 1e9 + 7;
const int inf = 2e18;
const int ninf = -2e18;
const int N = 1005;

int h,w;
int dp[N];
string grid;

int pow(int a, int b, int m)
{
	int ans = 1;
	while (b)
	{
		if (b & 1)
			ans = (ans * a) % m;
		b /= 2;
		a = (a * a) % m;
	}
	return ans;
}
bool inBound(int x){
	if(x >= 1 && x <= h*w){
		return true;
	}
	return false;
}
int solve(int s){
	if(dp[s] != -1) return dp[s];
	if(s == 1) return dp[1] = 1;
	int up = s - w;
	int left;
	if(s%w == 1) left = 1000005;
	else left = s - 1;
	int ans = 0;
	if(inBound(up)){
		if(grid[up] != '#')	ans  = ((ans%mod) + (solve(up)%mod))%mod;
		
	}
	if(inBound(left)){
		if(grid[left] != '#')	ans  = ((ans%mod) + (solve(left)%mod))%mod;
	}
	return dp[s] = ans;
}
signed main()
{

	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	#ifdef __APPLE__
		freopen("input.txt","r",stdin);
		freopen("output.txt","w",stdout);
		freopen("error.txt","w",stderr);
	#endif

	time_t t1, t2;
	t1 = clock();
	
	cin>>h>>w;
	grid = "Y";
	for(int i = 0; i < h ; i++){
		string inp;
		cin>>inp;
		grid += inp;
	}
	memset(dp,-1,sizeof(dp));
	
	cout<<solve(h*w)<<endl;
	
	t2 = clock();
	cerr<<endl<<t2-t1<<endl;
	
	return 0;
}