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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
void print_err() {
	cerr << "\n";
}
template <class T, class... Arg>
void print_err(T x, Arg&&... args) {
	cerr << x << " ";
	print_err(args...);
}
template<class T>
void print_container(T &cont) {
	for (auto iter : cont) {
		cerr << iter << " ";
	}
	cerr << "\n";
}
#ifdef local
#define debug(...) print_err(_VA_ARGS_)
#define debug_c(container) print_container(container)
#else
#define debug(...)
#define debug_c(container)
#endif
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define mod 1000000007
#define ll long long int
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define pll pair<ll,ll>
#define vl vector<ll>
#define vpll vector<pair<ll,ll>>
#define mll map<ll,ll>
#define mllrev map<ll,ll,greater<int>>
#define ps(x,y) fixed<<setprecision(y)<<x
#define w(t) ll t;cin>>t;while(t--)
#define fo(i,n) for(ll i=0;i<n;++i)
#define foab(i,a,b) for(ll i=a;i<b;++i)
#define inp cin>>
#define pr cout<<
#define all(A) (A).begin(),(A).end()
#define fastio ios_base::sync_with_stdio(NULL);cin.tie(NULL);cout.tie(NULL);
class Graph {
public:
	list<ll> *adj_list;
	ll v;
	Graph(ll v) {
		this->v = v;
		adj_list = new list<ll>[v];
	}
	void add_edge(ll u, ll v, bool bidirectional = true) {
		adj_list[u].push_back(v);
		if (bidirectional) {
			adj_list[v].push_back(u);
		}
	}
	void print();
	void connected_compo();
	void dfs(vector<vector<ll>> vect, ll k1, ll k2, ll n);
	void bfs(ll v, bool visited[]);
};
ll recursive(vector<vector<int>> &vect, vector<vector<int>> &dp, ll n, ll m, ll r, ll c) {
	if (r == 0 and c == 0) {
		return 1;
	}
	if (r < 0 or c < 0) {
		return 0;
	}
	if (vect[r][c] == -1) {
		return 0;
	}
	if (dp[r][c] != 0) {
		return dp[r][c];
	}
	return dp[r][c] = (recursive(vect, dp, n, m, r - 1, c) % mod + recursive(vect, dp, n, m, r, c - 1) % mod) % mod;
}
void solve() {
	ll n, m;
	inp n >> m;
	char ch;
	vector<vector<int>> vect(n, vector<int>(m, 0)), dp(n, vector<int>(m, 0));
	// memset(vect, 0, sizeof(vect));
	// memset(dp, 0, sizeof(dp));
	fo(i, n) {
		fo(j, m) {
			inp ch;
			if (ch == '#') {
				vect[i][j] = -1;
			}
		}
	}
	cout << recursive(vect, dp, n, m, n - 1, m - 1) << endl;
	// fo(i, n) {
	// 	fo(j, m) {
	// 		cout << vect[i][j] << " ";
	// 	}
	// 	cout << endl;
	// }
	return;
}
//Rod cutting problem test case
// 5
// 100
// 23 15 50
// 4000
// 3 4 5
// 7
// 5 5 2
// 4
// 2 1 1
// 5
// 5 3 2
int main() {
	fastio
#ifdef local
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	freopen("error.txt", "w", stderr);
#endif
	// w(t) {
	solve();
	// }
	return 0;
}