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: Runtime Error)

#include <bits/stdc++.h>
typedef long long int ll;
using namespace std;
#define mod 1000000007
vector<ll> A,tree(4*1e5, 0);
ll n,a,b,k,h,m;
void build(int node, int start, int end,int p)
{
    if(start == end)
    {
        // Leaf node will have a single element
        tree[node] = A[start];
    }
    else
    {
        int mid = (start + end) / 2;
        // Recurse on the left child
        build(2*node, start, mid,p+1);
        // Recurse on the right child
        build(2*node+1, mid+1, end,p+1);
        // Internal node will have the sum of both of its children
        if(p%2==n%2)
        {
             tree[node] = tree[2*node] | tree[2*node+1];
        }
        else
        {
             tree[node] = tree[2*node] ^ tree[2*node+1];
        }



         }
    }

void update(int node, int start, int end, int idx, int val,int p)
{
    if(start == end)
    {
        // Leaf node
        A[idx] = val;
        tree[node] = val;
    }
    else
    {
        int mid = (start + end) / 2;
        if(start <= idx and idx <= mid)
        {
            // If idx is in the left child, recurse on the left child
            update(2*node, start, mid, idx, val,p+1);
        }
        else
        {
            // if idx is in the right child, recurse on the right child
            update(2*node+1, mid+1, end, idx, val,p+1);
        }
        // Internal node will have the sum of both of its children
        if(p%2==n%2)
        {
             tree[node] = tree[2*node] | tree[2*node+1];
        }
        else
        {
             tree[node] = tree[2*node] ^ tree[2*node+1];
        }
    }
}
int main(){

        cin>>n>>m;

        for(ll i=0;i<pow(2,n);i++)
        {
            cin>>a;
            A.push_back(a);
        }
        build(1,0,A.size()-1,1);
        for(ll i=0;i<m;i++)
        {
            cin>>a;
            cin>>b;
            update(1,0,A.size()-1,a-1,b,1);
            cout<<tree[1]<<endl;
        }

    return 0;

    }