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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>   //PBDS
#include <ext/pb_ds/tree_policy.hpp>       //PBDS
#define int long long int
#define vi vector<int>
#define vii vector<pair<int,int>>
#define pi pair<int,int>
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define mi map<int,int>
#define PI   3.141592653589
#define all(a)  (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define f(i,x,n) for(int i=x;i<n;i++)
#define rf(i,x,n) for(int i=n-1;i>=x;i--)
#define check(x)   cerr<<#x<<" : "<<x<<endl;
#define ordered_set tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update>
#define fbo find_by_order                    //return iterator PBDS
#define ook order_of_key                     //return number :)) PBDS
#define baar exit(0)
#define INF 1000000000000000000
using namespace __gnu_pbds;
using namespace std;
int powmod(int base,int exponent,int mod)
{
	int ans=1;
	while(exponent){
		if(exponent&1)ans=(ans*base)%mod;
		base=(base*base)%mod;
		exponent/=2;
	}
	return ans;
}
signed main()
{
	fast
	int n,m;
	cin>>n>>m;
	int mod=1000000007;
	string s[n];
	f(i,0,n)
	{
		cin>>s[i];
	}
	int dp[n][m];
	if(s[0][0]=='.') dp[0][0]=1;
	else dp[0][0]=0;
	f(i,0,n)
	{
		f(j,0,m)
		{
			if(i==0 && j==0) continue;
			if(s[i][j]=='#')
			{
				dp[i][j]=0;
				continue;
			}
			if(i==0)
			{ 
			    dp[i][j]=dp[i][j-1];
			}
			else if(j==0)
			{
			    dp[i][j]=dp[i-1][j];
			}
			else
			{
				dp[i][j]=(dp[i-1][j]+dp[i][j-1])%mod;
			}
		}
	}
	cout<<dp[n-1][m-1]%mod;
	return 0;
}