#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

typedef long long ll;

const int maxn = ;

const int pow[] = {,,,,,,,,,,,,};

int n,m,now,pre,tx,ty;

char map[][];

struct node

{

    int head[maxn],nextt[maxn],tot;

    ll sum[maxn],sta[maxn];

    void clear()

    {

        memset(head,-,sizeof(head));

        tot = ;

    }

    void push(ll x,ll v)

    {

        ll hashh = x % maxn;

        for (int i = head[hashh]; i >= ; i = nextt[i])

        {

            if (sta[i] == x)

            {

                sum[i] += v;

                return;

            }

        }

        sta[tot] = x;

        sum[tot] = v;

        nextt[tot] = head[hashh];

        head[hashh] = tot++;

    }

} f[];

int turnleft(ll x,int k)

{

    return x << pow[k];

}

int get(ll x,int k)

{

    return (x >> pow[k]) & ;

}

ll del(ll x,int i,int j)

{

    return x & (~( << pow[i])) & (~( << pow[j]));

}

int findr(ll x,int pos)

{

    int cnt = ;

    for (int i = pos + ; i <= m; i++)

    {

        int k = get(x,i);

        if (k == )

            cnt++;

        else if (k == )

            cnt--;

        if (!cnt)

            return i;

    }

}

int findl(ll x,int pos)

{

    int cnt = ;

    for (int i = pos - ; i >= ; i--)

    {

        int k = get(x,i);

        if (k == )

            cnt++;

        else if (k == )

            cnt--;

        if (!cnt)

            return i;

    }

}

void solve2(int x,int y,int k)

{

    int p = get(f[pre].sta[k],y - );   //右插头

    int q = get(f[pre].sta[k],y);   //下插头

    ll staa = del(f[pre].sta[k],y - ,y);   //将这两个插头删掉以后的状态

    ll v = f[pre].sum[k];

    if (!p && !q)  //新建一个连通分量

    {

        if (map[x][y] == '*')

        {

            f[now].push(staa,v);

            return;

        }

        if (x < n && y < n && map[x + ][y] == '.' && map[x][y + ] == '.')

            f[now].push(staa | turnleft(,y - ) | turnleft(,y),v);

    }

    else if (!p || !q)  //保持原来的连通分量

    {

        int temp = p + q;

        if (x < n && map[x + ][y] == '.')

            f[now].push(staa | turnleft(temp,y - ),v);

        if (y < m && map[x][y + ] == '.')

            f[now].push(staa | turnleft(temp,y),v);

    }

    else if (p ==  && q == )    //连接两个联通分量

        f[now].push(staa ^ turnleft(,findr(staa,y)),v);  //这里的异或实际上就是把1变成2,2变成1

    else if (p ==  && q == )

        f[now].push(staa ^ turnleft(,findl(staa,y - )),v);

    else if (p ==  && q == )

        f[now].push(staa,v);

    else if (x == tx && y == ty)

        f[now].push(staa,v);

}

ll solve()

{

    f[].clear();

    f[].push(,);

    now = ,pre = ;  //滚动数组

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

    {

        pre = now;

        now ^= ;

        f[now].clear();

        for (int k = ; k < f[pre].tot; k++)

            f[now].push(turnleft(f[pre].sta[k],),f[pre].sum[k]);   //左移一位，因为轮廓线下来的时候会少一个插头

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

        {

            pre = now;

            now ^= ;

            f[now].clear();

            for (int k = ; k < f[pre].tot; k++)

                solve2(i,j,k);  //处理第k个状态

        }

    }

    for (int i = ; i < f[now].tot; i++)

        if (f[now].sta[i] == )  //没有插头了.

            return f[now].sum[i];

    return ;

}

int main()

{

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

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

        scanf("%s",map[i] + );

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

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

            if (map[i][j] == '.')

                tx = i,ty = j; //找右下角的非障碍点

    if (!tx)

        puts("");

    else

        printf("%lld\n",solve());

    return ;

}