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>
#include<ext/pb_ds/tree_policy.hpp>

// find_by_order(k) returns the iterator to the kth largest element (start from 0)
// order_of_key(k) returns the number of elements in the set which are strictly less than the value k

using namespace std;
using namespace __gnu_pbds;

#define ff              first
#define ss              second
#define all(v)          v.begin(), v.end()
#define int             long long
#define pb              push_back
#define pii             pair<int,int>
#define vi              vector<int>
#define mem(dp,x)       memset(dp,x,sizeof dp);
#define mii             map<int,int>
#define pqb             priority_queue<int>
#define pqs             priority_queue<int,vi,greater<int> >
#define setbits(x)      __builtin_popcountll(x)
#define rightmostbit(x) __builtin_ffsll(x)
#define mod             1000000007
#define inf             1e18
#define pi              3.141592653589793238
#define mk(a,n)         int n;cin>>n;vi a(n);for(int i=0;i<n;i++) cin>>a[i]
#define w(x)            int x; cin>>x; while(x--)
#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }

void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
	cerr << *it << " = " << a << endl;
	err(++it, args...);
}

int dx[] = { -1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
int dx2[] = { -1 , -1 , -1, 0 , 0 , 1 , 1 , 1};
int dy2[] = { -1 , 0 , 1 , -1, 1 , -1, 0, 1};

typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;

void g_b()
{
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	freopen("error.txt", "w", stderr);
#endif
}
// .Always initialize dp array
// 1, 1, 2, 5, 14, 42, 132, 429, 1430, CATALAN NUMBER
int32_t main()
{
	g_b();
	int n, m;
	cin >> n >> m;
	char a[n][m];
	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++)
			cin >> a[i][j];
	int dp[n][m];
	bool f = 0, f1 = 0;
	if (a[n - 1][m - 1] == '#')
	{
		cout << 0;
		return 0;
	}
	memset(dp, 0, sizeof dp);
	for (int i = n - 1; i >= 0; i--)
	{
		for (int j = m - 1; j >= 0; j--)
		{
			if (i == n - 1)
			{
				if (a[i][j] == '#')
				{
					dp[i][j] = 0;
					f = 1;
				}
				else if (!f)
					dp[i][j] = 1;
			}
			else if (j == m - 1)
			{
				if (a[i][j] == '#')
				{
					dp[i][j] = 0;
					f1 = 1;
				}
				else if (!f1)
					dp[i][j] = 1;
			}
			else
			{
				if (a[i][j] == '#')
					dp[i][j] = 0;
				else
					dp[i][j] = (dp[i + 1][j] + dp[i][j + 1]) % mod;
			}
		}
	}

	cout << dp[0][0] % mod;
	return 0;

}