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 fo(i, n) for(ll i = 0; i < n; i++)
#define endl "\n"
#define of(i, n) for(ll i = n - 1; i >= 0; i--)
#define ll long long
#define vec vector<ll>
#define fio ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define tr(container, it) \
for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)


ll mod= 1e9+7;

#define umap unordered_map

string sp,tp;


/*string st(string s, string t)
{
    ll n = s.length();
    ll m = t.length();
    if(s.length()==0 || t.length()==0) {dp[n][m]=0;return "";}
    if(dp[s.length()][t.length()]!=0) return  st(s,t);
    else if(s[0]==t[0])
    {
        dp[n][m]++;
        s.erase(s.begin());
        t.erase(t.begin());
        return st(s,t);
    }
}*/

ll ncr(ll n, ll r)
{
    long long p = 1, k = 1;
    if (n - r < r)
        r = n - r;

    if (r != 0) {
        while (r) {
            p *= n;
            k *= r;
            long long m = __gcd(p, k);
            p /= m;
            k /= m;

            n--;
            r--;
        }
    }

    else
        p = 1;
    return p;
}
ll dp[1001][1001];
int main() {
    fio;
    ll n,m;
    cin>>n>>m;
    char c[n+1][m+1];
    for (int i=1; i<=n; i++){
		for (int j=1; j<=m; j++) cin >> c[i][j];
	}
    ll sum=0;
    dp[1][1]=1;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(c[i][j]=='.') dp[i][j] += dp[i][j-1] + dp[i-1][j];
            dp[i][j] = dp[i][j] % mod;
        }
    }
    cout<<dp[n][m];
}