1214_D. Treasure Island_21609
All of us love treasures, right? That's why young Vasya is heading for a Treasure Island.

Treasure Island may be represented as a rectangular table n × m which is surrounded by the ocean. Let us number rows of the field with consecutive integers from 1 to n from top to bottom and columns with consecutive integers from 1 to m from left to right. Denote the cell in r-th row and c-th column as (r, c). Some of the island cells contain impassable forests, and some cells are free and passable. Treasure is hidden in cell (n, m).

Vasya got off the ship in cell (1, 1). Now he wants to reach the treasure. He is hurrying up, so he can move only from cell to the cell in next row (downwards) or next column (rightwards), i.e. from cell (x, y) he can move only to cells (x+1, y) and (x, y+1). Of course Vasya can't move through cells with impassable forests.

Evil Witch is aware of Vasya's journey and she is going to prevent him from reaching the treasure. Before Vasya's first move she is able to grow using her evil magic impassable forests in previously free cells. Witch is able to grow a forest in any number of any free cells except cells (1, 1) where Vasya got off his ship and (n, m) where the treasure is hidden.

Help Evil Witch by finding out the minimum number of cells she has to turn into impassable forests so that Vasya is no longer able to reach the treasure.

Input

First line of input contains two positive integers n, m (3 ≤ n ⋅ m ≤ 1 000 000), sizes of the island.

Following n lines contains strings s_i of length m describing the island, j-th character of string s_i equals "#" if cell (i, j) contains an impassable forest and "." if the cell is free and passable. Let us remind you that Vasya gets of his ship at the cell (1, 1), i.e. the first cell of the first row, and he wants to reach cell (n, m), i.e. the last cell of the last row.

It's guaranteed, that cells (1, 1) and (n, m) are empty.

Output

Print the only integer k, which is the minimum number of cells Evil Witch has to turn into impassable forest in order to prevent Vasya from reaching the treasure.

Examples

Input


2 2
..
..


Output


2


Input


4 4
....
#.#.
....
.#..


Output


1


Input


3 4
....
.##.
....


Output


2

Note

The following picture illustrates the island in the third example. Blue arrows show possible paths Vasya may use to go from (1, 1) to (n, m). Red illustrates one possible set of cells for the Witch to turn into impassable forest to make Vasya's trip from (1, 1) to (n, m) impossible.

<image>
from __future__ import division, print_function
def main():
    n,m=map(int,input().split())
    grid=[]
    for i in range(n):
        grid.append(list(input()))
    def dfs(temp):
        visited=[0]*(m*n)
        stack=[0]
        ans=0
        while stack:
            s=stack.pop()
            x,y=s//m,s%m
            visited[x*m+y]=1
            if y+1<m:
                if visited[x*m+(y+1)]==0 and temp[x][y+1]=='.':
                    stack.append(x*m+(y+1))
                    if x*m+y+1==n*m-1:
                        visited[n*m-1]=1
                        break
            if x+1<n:
                if visited[(x+1)*m+y]==0 and temp[x+1][y]=='.':
                    stack.append((x+1)*m+y)
                    if (x+1)*m+y==n*m-1:
                        visited[-1]=1
                        break
        if visited[-1]==1:
            ans+=1
        visited[0]=0
        visited[-1]=0
        stack=[0]
        while stack:
            s=stack.pop()
            x,y=s//m,s%m
            visited[s]=1
            
            if x+1<n:
                if visited[(x+1)*m+y]==0 and temp[x+1][y]=='.':
                    stack.append((x+1)*m+y)
                    visited[(x+1)*m+y]=1
            if y+1<m:
                if visited[x*m+y+1]==0 and temp[x][y+1]=='.':
                    stack.append(x*m+y+1)
                    visited[x*m+y+1]=1
        if visited[-1]==1:
            ans+=1
        print(ans)
        
    
    dfs(grid)
 
        
    
 
 
######## Python 2 and 3 footer by Pajenegod and c1729
 
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.
 
# So on cf, use PyPy2 for best string performance.
 
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
 
import os, sys
from io import IOBase, BytesIO
 
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
 
# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'0' [0]:
        A.append(sign*numb)
    return A
 
if __name__== "__main__":
  main()