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)

// ░░░░░░░░( •̪●)░░░░░░░░░░░░░░░░░░░░░░░░
// ░░░░░░███████ ]▄▄▄▄▄▄▄▄▃░░░▃░░░░ ▃░
// ▂▄▅█████████▅▄▃▂░░░░░░░░░░░░░░░░░
// [███████████████████].░░░░░░░░░░░░░░
// ◥⊙▲⊙▲⊙▲⊙▲⊙▲⊙▲⊙◤...░░░░░░░░░░░░

//PointBlank's code (⌐■_■)
 
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
 
#define repp(i, a, b) for(ll i = a; i <= b; i++)
#define rep(i, a, b) for(ll i = a; i < b; i++)
#define rrep(i, b, a) for(ll i = b; i >= a; i--)
#define pb emplace_back
#define ff first
#define ss second
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define lb lower_bound
#define ub upper_bound
#define mp make_pair
#define mem(a, b) memset(a, b, sizeof(a));
#define mem0(a) memset(a, 0, sizeof(a));
#define quickio ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define in_edges(m) repp(i, 1, m){ll a, b; cin >> a >> b; v[a].pb(b), v[b].pb(a);}
#define debug(this) cerr<<"> "<<#this<<" : "<<this<<"\n"
#define endl "\n"
ll power(ll x, ll y); 

const ll MOD = 1000000007;
const ll INF = 1e18L + 5;
ll res[1003][1003];

int main() //PointBlank's code ¯\_(ツ)_/¯
{
    quickio
    ll n, m;
    cin >> n >> m;
    char mat[n][m];
    rep(i, 0, n) rep(j, 0, m) cin >> mat[i][j];
    if(mat[0][0] != '#') res[1][0] = 1;
    
    repp(i, 1, n) repp(j, 1, m) if(mat[i - 1][j - 1] != '#') res[i][j] = (res[i][j - 1] + res[i - 1][j])%MOD;
    cout << res[n][m];
}