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;
typedef long long lli;
typedef long double lld;
#define mod 1000000007
#define mod2 998244353
#define fori(n) for (int i = 0; i < n; i++)
#define forj(n) for (int j = 0; j < n; j++)
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define fast ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define endl "\n"
#define sortvec(a) sort(a.begin(), a.end())
#define watch(x) cerr << "\n" << (#x) << " is " << (x) << endl
#define cerr if(true)cerr
void print(){cout<<endl;}
template<typename T,typename... Args>
void print(T a, Args... args) {cout<<a<<" ";print(args...);}
lli llipowerp(lli x, lli y, lli p=LLONG_MAX)
{
    lli res = 1;
    x = x % p;
    while (y > 0)
    {
        if (y & 1) res = (res * x) % p;
        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}
lli binarySearch(vector<int> arr, int l, int r, int x)
{
    if (r >= l)
    {
        lli mid = l + (r - l) / 2;
        if (arr[mid] == x)
            return mid;
        if (arr[mid] > x)
            return binarySearch(arr, l, mid - 1, x);
        return binarySearch(arr, mid + 1, r, x);
    }
    return -1;
}
template <typename T>
inline void printvec(vector<T> inp){
    for(auto it:inp) cout<<it<<" ";
    cout << endl;
}
//---------------------------------------------------------------------------------
int h,w,dp[1010][1010];
vector<string> a(1010);


int fill(int i,int j){
    if(i<=0 || j<=0) return 0;
    if(dp[i][j]!=-1) return dp[i][j];
    if(a[i][j]=='#'){
        dp[i][j]=0;
        return 0;
    }

    int ans=0;
    if(a[i-1][j]=='.') ans=(ans+fill(i-1,j))%mod;
    if (a[i][j-1] == '.')
        ans = (ans + fill(i, j-1)) % mod;
    dp[i][j]=ans;
    return ans;
}

int main()
{   
    cin>>h>>w;
    memset(dp,-1,sizeof(dp));
    string x;
    fori(w) x+='#';
    a[0]=x;
    dp[1][1]=1;
    fori(h){
        cin>>x;
        a[i+1]='#'+x;

    }
    cout<<fill(h,w)<<endl;

    
}