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: Wrong Answer)

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

typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef pair<ll,ll> P;
typedef vector<P> vp;

const ll mod = 1e9+7;
const ll inf = 1e16;

#define loop(i,m,n) for(ll i=m;i<n;i++)
#define vloop(it,x) for(auto it=x.begin(); it!=x.begin(); it++)
#define all(x) x.begin(),x.end()
#define debug(x) cout << #x <<":"<< x <<endl	
#define showll(x) printf("%lld\n",x)
#define test(t) ll t;cin >> t;while(t--)
#define showc(x) printf("%c\n",x)
#define show(x) cout << x << endl
#define read(x) cin >> x
#define yes printf("YES\n")
#define no printf("NO\n")
#define pb push_back
#define mp make_pair

void fast()
{
ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
}

#define MAXN 60000001
int spf[MAXN];//smallest prime factor 
  
void sieve() 
{ 
    spf[1] = 1; 
    for (int i=2; i<MAXN; i++) 
    {
        spf[i] = i; 
    }
   
    for (int i=4; i<MAXN; i+=2) 
    {
        spf[i] = 2; 
    }
  
    for (int i=3; i*i<MAXN; i++) 
    {  
        if (spf[i] == i) 
        {  
            for (int j=i*i; j<MAXN; j+=i) 
  			{
                if (spf[j]==j) 
                {	
                	spf[j] = i;
                }
            } 
        } 
    } 
}

//////////////////////////////// - _ - \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ 
  
const int maxi = 1005;
char sl[maxi][maxi];

int main()
{
	ll h,w,i,j,a,b;
	ll ans = 0;
	read(h);read(w);
	loop(i,0,h)
	{
		scanf("%s",sl[i]);
	}
	vector< vector <ll> > dp(h, vector <ll> (w));
	dp[0][0]=1;
	loop(i,0,h)
	{
		loop(j,0,w)
		{
			if(i+1<h && sl[i+1][j]=='.')
			{
				dp[i+1][j] = dp[i+1][j] + dp[i][j]; 
			}
			if(j+1<w && sl[i][j+1]=='.')
			{
				dp[i][j+1] = dp[i][j+1] + dp[i][j];	
			}
		}
	}
	

	printf("%lld\n",dp[h-1][w-1]);

}