// Original C++ solution (slower):
#include<bits/stdc++.h>

#define ll long long

#define ull unsigned long long

#define IO ios::sync_with_stdio(false), cin.tie(0)

#define endl '\n'

#define main() int main()

#define ff1(i,n) for(int i=1;i<=n;++i)

#define ff0(i,n) for(int i=0;i<n;++i)

#define pi acos(-1.0)

#define ms(a, x) memset(a, x, sizeof(a))

#define lowbit(x) ((x)&(-(x)))

#define pb push_back

#define binarycount __builtin_popcount(x)

#define tr t[root]

#define lson t[root << 1]

#define rson t[root << 1 | 1]

#define fir first

#define sec second

#define debug(x) cout<<x<<'\n'

#define sa(a) scanf("%d", &a)

#define sab(a, b) scanf("%d%d", &a, &b)

#define pa(a) printf("%d",a)

#define pab(a,b) printf("%d%d",a,b)

#define pii pair<string,int>

using namespace std;

#define maxn 1005

const int inf = 0x3f3f3f3f;

const int mod = 1e9+7;

const double eps = 1e-8;

ll gcd(ll a, ll b) {return b==0?a:gcd(b, a%b);}

ll lcm(ll a, ll b) {return a/gcd(a, b)*b;}

int max(int x,int y){if(x>y) return x;return y;}

ll powmod(ll a,ll b){

    ull ans = 1, pow = a%mod;

    while(b) { if(b&1) ans = (ans*pow)%mod; pow = (pow*pow)%mod; b >>= 1; }

    return ans%mod;

}



int n,m;

char mp[maxn][maxn];

int dp[maxn][maxn];

main()

{

	IO;

	cin>>n>>m;

	ms(dp,0);

	for(int i=1;i<=n;i++)

		for(int j=1;j<=m;j++)

		 cin>>mp[i][j];

		 

	for(int i=1;i<=n;i++){

		if(mp[i][1]=='#')break;

		dp[i][1]=1;

	}	 

	for(int i=1;i<=m;i++){

		if(mp[1][i]=='#')break;

		dp[1][i]=1;

	}

	for(int i=2;i<=n;i++)

	{

		for(int j=2;j<=m;j++)

		{

			if(mp[i][j]=='#')continue;

			if(mp[i-1][j]!='#'){

				dp[i][j]=(dp[i][j]+dp[i-1][j])%mod;

			}

			if(mp[i][j-1]!='#'){

				dp[i][j]=(dp[i][j]+dp[i][j-1])%mod;

			}

		}

	}



	cout<<dp[n][m]%mod<<endl;

	

}


// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f

#define INF 0x3f3f3f3f

#define pb push_back

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

#define ms(a, x) memset(a, x, sizeof(a))

#define for0(i, n) for(int i = 0; i < n; ++i)

#define for1(i, n) for(int i = 1; i <= n; ++i)

using namespace std;

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int, int> pii;

typedef pair<long long, int> pli;

typedef pair<long long, long long> pll;

const int mod = 1e9 + 7;

const int N = 1e3+10;

const int M = 1e6+10;

inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

inline ll Pow(ll a, ll b) { ll ans = 1; for( ; b; b >>= 1) { if(b&1) ans = ans * a % mod; a = a * a % mod;} return ans;}

inline ll Mul ( ll a, ll b)  { ll lf = a * ( b >> 25ll ) % mod * ( 1LL << 25 ) % mod; ll rg = a * ( b & ( ( 1ll << 25 ) - 1 ) ) % mod; return ( lf + rg ) % mod;}

inline ll C(ll y, ll x) { ll ans = 1; if(y < 0 || x < 0 || y < x) return 0; y %= mod; if(!y || !x) return 1; for0(i, x) ans = ans * (y-i) % mod; for1(i, x) ans = ans * Pow(i, mod-2) % mod; return ans; }

//==================================================================================================================================================================//



int n, m, dp[N][N];

char mp[N][N];



int main() {

   dp[1][1] = 1;

   scanf("%d %d", &n, &m);

   for1(i, n) for1(j, m) {

      scanf(" %c", &mp[i][j]);

   }

   for1(i, n) for1(j, m) {

      if(mp[i][j] == '#' || (i==1 && j==1)) continue;

      dp[i][j] = (dp[i-1][j] + dp[i][j-1]) % mod;

   }

   printf("%d\n", dp[n][m]);

   return 0;

}

// Fastest known solution for this problem: