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 5.4.1) (judge verdict: Runtime Error)

/*
	Powered by C++11.
	Author : Alex_Wei.
*/

#include <bits/stdc++.h>
using namespace std;

//#pragma GCC optimize(3)
//#define int long long

#define ll long long
#define pii pair <int,int>
#define fi first
#define se second
#define pb emplace_back
#define all(x) x.begin(),x.end()
#define sor(x) sort(all(x))
#define rev(x) reverse(all(x))
#define mem(x,v) memset(x,v,sizeof(x))

// Templates start.

namespace IO{
	char buf[1<<21],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;
	#ifdef __WIN64
		#define gc getchar()
	#else
		#define gc (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin),p1==p2)?EOF:*p1++)
	#endif
	#define pc(x) (*O++=x)
	#define flush() fwrite(obuf,O-obuf,1,stdout)

	inline int read(){
		int x=0,sign=0; char s=getchar();
		while(!isdigit(s))sign|=s=='-',s=getchar();
		while(isdigit(s))x=(x<<1)+(x<<3)+(s-'0'),s=getchar();
		return sign?-x:x;
	}
	void print(int x) {if(x>9)print(x/10); pc(x%10+'0');}
}

namespace math{
	const int mod=1e9+7;
	const int maxn=1e6+5;

	ll ksm(ll a,ll b){ll s=1,m=a; while(b){if(b&1)s=s*m%mod; m=m*m%mod,b>>=1;} return s;}
	ll inv(ll x){return ksm(x,mod-2);}

	ll fc[maxn],ifc[maxn];
	void init_m(int n){
		fc[0]=1; for(int i=1;i<=n;i++)fc[i]=fc[i-1]*i%mod;
		ifc[n]=inv(fc[n]); for(int i=n-1;i>=0;i--)ifc[i]=ifc[i+1]*(i+1)%mod;
	}
	ll C(ll n,ll m){return fc[n]*ifc[m]%mod*ifc[n-m]%mod;}
}

// Templates end. Please give me more points.

using namespace IO;
using namespace math;

const int N=3e3+5;

int h,w,n,f[N],ans;
struct point{
	int x,y;
	bool operator < (const point &v) const {
		return x==v.x?y<v.y:x<v.x;
	}
}c[N];

int main(){
	cin>>h>>w; init_m(h+w); f[0]=0;
	for(int i=1;i<=h;i++)for(int j=1;j<=w;j++){char p; cin>>p; if(p=='#'||i==h&&j==w)c[++n]={i,j};}
	for(int i=0;i<=n;i++){
		f[i]=C(c[i].x+c[i].y-2,c[i].x-1);
		for(int j=0;j<i;j++){
			int a=c[i].x-c[j].x,b=c[i].y-c[j].y;
			if(a>=0&&b>=0)f[i]=(f[i]-1ll*C(a+b,a)*f[j]%mod+mod)%mod;
		}
		if(i<n&&i)ans=(ans+1ll*f[i]*C(h+w-c[i].x-c[i].y,h-c[i].x))%mod;
	}
	cout<<(C(h+w-2,h-1)%mod-ans+mod)%mod<<endl;
	return 0;
}