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 FOR(I,X,Y) for(long long (I)=(X);(I)<(Y);(I)++)
#define REP(I,X,Y) for(long long (I)=(Y)-1;(I)>=(X);(I)--)
#define ALL(X) (X).begin(),(X).end()
#define pb push_back
#define COUNT(V,X) upper_bound((V).begin(),(V).end(),X)-lower_bound((V).begin(),(V).end(),X)
#define debug(x) cerr<<#x<<':'<<x<<endl;
#define DEBUG(v) cerr<<#v<<':';for(auto xXx:v)cerr<<xXx<<' ';cerr<<endl;
#define INF 1000000007
#define LINF 1000000000000000007
#define Yes(X) if(X){cout<<"Yes"<<endl;}else{cout<<"No"<<endl;}
#define YES(X) if(X){cout<<"YES"<<endl;}else{cout<<"NO"<<endl;}
#define ctoi(C) (C-'0')
typedef long long ll;
typedef long double ld;
long long dx[] = {1,0,-1,0};
long long dy[] = {0,1,0,-1};
long long dx8[] = {1,1,0,-1,-1,-1,0,1};
long long dy8[] = {0,1,1,1,0,-1,-1,-1};
long long dx9[] = {1,1,0,-1,-1,-1,0,1,0};

signed main(){
    ll H,W;
    cin >> H >> W;
    char a[H][W];
    FOR(i,0,H)FOR(j,0,W)cin >> a[i][j];
    ll DP[H][W];
    FOR(i,0,H)FOR(j,0,W)DP[i][j] = 0;
    FOR(i,0,H)FOR(j,0,W){
        if(a[i][j] == '#'){
            DP[i][j] = 0;
        }
        else if(i == 0 && j == 0){
            DP[i][j] = 1;
        }
        else if(i == 0){
            DP[i][j] = DP[i][j-1];
        }
        else if(j == 0){
            DP[i][j] = DP[i-1][j];
        }
        else{
            DP[i][j] = DP[i-1][j]+DP[i][j-1];
        }
        DP[i][j] %= INF;
    }
    cout << DP[H-1][W-1] << endl;
}