// Original C++ solution (slower):


/**

    And in the end, the love you take is equal to the love you make.



        I LOVE YOU LAURA.

**/



#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <ext/rope>



#define _CRT_SECURE_NO_DEPRECATE

#define _CRT_DISABLE_PERFCRIT_LOCKS



#define findOrder find_by_order

#define orderKey order_of_key



#define isOn(mask, i) (mask & (1 << i))

#define turn(mask, i) (mask |= (1 << i))

#define switch(mask, i) (mask &= ~(1 << i))

#define change(mask, i) (mask ^= (1 << i))

#define lowbit(mask) (mask & (-mask))

#define turnAll(mask, n) (mask = (1 << n) - 1)

#define switchLastBit(mask) (mask & (mask - 1))

#define turnLastZero(mask) (mask | (mask + 1))

#define modulo(x, mod) ((x) & (mod - 1))

#define log2(x) (31 - __builtin_clz(x))



#define mkp make_pair

#define pb push_back

#define pf push_front

#define eb emplace_back

#define ef emplace_front

#define ins insert

#define rsz resize

#define f first

#define s second

#define lb lower_bound

#define ub upper_bound



#define sz(x) ((int)x.size())

#define all(x) begin(x), end(x)

#define rall(x) rbegin(x), rend(x)



#define debug(x) (cerr << x << " ")



using namespace std;

using namespace __gnu_pbds;

using namespace __gnu_cxx;



typedef double db;

typedef long long ll;

typedef long double ld;

typedef unsigned long long ull;



typedef pair<int, int> ii;

typedef pair<ll,ll> pll;

typedef pair<ld,ld> pdb;

typedef pair<int, ii> tri;

typedef complex<ld> cd;



typedef vector<int> vi;

typedef vector<ii> vii;

typedef vector<ld> vdb;

typedef vector<cd> vcd;

typedef vector<ll> vll;

typedef vector<string> vs;

typedef vector<vll> matrix;



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



const ll mod = 1e9+7;

const int maxn = 1005;

const int pi = 2 * acos(0.0);

const db eps = 1e-9;



int n, m;

char arr[maxn][maxn];



bool ok (int x, int y){

    return (x >= 1 && x <= n && y >= 1 && y <= m);

}



int dx[] = {0, 1};

int dy[] = {1, 0};



ll dp[maxn][maxn];



ll dfs (int x, int y){

    if (!ok(x, y) || arr[x][y] == '#') return 0LL;

    if (x == n && y == m){

        return 1LL;

    }



    ll &ans = dp[x][y];



    if (ans != -1) return ans % mod;



    ans = 0;

    for (int i = 0; i < 2; i++){

        int nx = x + dx[i];

        int ny = y + dy[i];

        ans = (ans + dfs(nx, ny)) % mod;

    }



    return ans % mod;

}



int main()

{

    ios_base::sync_with_stdio(0);

    cin.tie(0); cout.tie(0);



    memset(dp, -1, sizeof dp);



    cin >> n >> m;



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

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

            cin >> arr[i][j];

        }

    }



    cout << dfs(1, 1) % mod << "\n";



    return 0;

}


// 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: