32_D. Constellation_5223
A star map in Berland is a checked field n × m squares. In each square there is or there is not a star. The favourite constellation of all Berland's astronomers is the constellation of the Cross. This constellation can be formed by any 5 stars so, that for some integer x (radius of the constellation) the following is true: 

  * the 2nd is on the same vertical line as the 1st, but x squares up 
  * the 3rd is on the same vertical line as the 1st, but x squares down 
  * the 4th is on the same horizontal line as the 1st, but x squares left 
  * the 5th is on the same horizontal line as the 1st, but x squares right 



Such constellations can be very numerous, that's why they are numbered with integers from 1 on the following principle: when two constellations are compared, the one with a smaller radius gets a smaller index; if their radii are equal — the one, whose central star if higher than the central star of the other one; if their central stars are at the same level — the one, whose central star is to the left of the central star of the other one.

Your task is to find the constellation with index k by the given Berland's star map.

Input

The first line contains three integers n, m and k (1 ≤ n, m ≤ 300, 1 ≤ k ≤ 3·107) — height and width of the map and index of the required constellation respectively. The upper-left corner has coordinates (1, 1), and the lower-right — (n, m). Then there follow n lines, m characters each — description of the map. j-th character in i-th line is «*», if there is a star in the corresponding square, and «.» if this square is empty.

Output

If the number of the constellations is less than k, output -1. Otherwise output 5 lines, two integers each — coordinates of the required constellation. Output the stars in the following order: central, upper, lower, left, right.

Examples

Input

5 6 1
....*.
...***
....*.
..*...
.***..


Output

2 5
1 5
3 5
2 4
2 6


Input

5 6 2
....*.
...***
....*.
..*...
.***..


Output

-1


Input

7 7 2
...*...
.......
...*...
*.***.*
...*...
.......
...*...


Output

4 4
1 4
7 4
4 1
4 7


stdin_flag=1
if not stdin_flag:
  read_line_index=0
  testcasefilename="test.txt"
  Stestcase=open(testcasefilename).read()
  read_line_datas=Stestcase.split("\n")


#####################################
######################################

def read_line():
  global read_line_index
  if stdin_flag:
    return raw_input()
  else:
    s=read_line_datas[read_line_index]
    read_line_index+=1
    return s

def answer():
  if stdin_flag:
    return solve()
  else:
    while read_line_proceed():
      solve()
      

def read_line_proceed():
  global read_line_index
  print"##################"
  while 1:
    if read_line_index>= len (read_line_datas ):
      return False
    if read_line_datas[read_line_index]=="%%%":
      read_line_index+=1
      return True
    read_line_index+=1


def readint():
  return int (read_line() )


def readints():
  return map(int, read_line().split(" "))

def reads():
  return read_line()

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def compute(n,m,a,k):
  s=0
  for r in xrange(1,301):
    for y in xrange(r,n-r):
      for x in xrange(r,m-r):
        if a[y][x] and a[y+r][x] and a[y-r][x] and a[y][x+r] and a[y][x-r]:
          s+=1
          if s==k:
            put(y,x,r)
            return

  print -1

def put(y0,x0,r):
  y=y0+1
  x=x0+1
  print y,x
  print y-r,x
  print y+r,x
  print y,x-r
  print y,x+r

def m2n(s):
  return [int(i=="*") for i in s]

def solve():
  n,m,k=readints()
  a=[m2n(reads()) for _ in xrange(n) ]
  compute(n,m,a,k)

def test():
  n=300
  compute(n,n,[[0 for i in xrange(n)]for j in xrange(n)],10**8)
  pass

#test()
answer()
