There is a rectangular grid of *n* rows of *m* initially-white cells each.

Arkady performed a certain number (possibly zero) of operations on it. In the *i*-th operation, a non-empty subset of rows *R*<sub class="lower-index">*i*</sub> and a non-empty subset of columns *C*<sub class="lower-index">*i*</sub> are chosen. For each row *r* in *R*<sub class="lower-index">*i*</sub> and each column *c* in *C*<sub class="lower-index">*i*</sub>, the intersection of row *r* and column *c* is coloured black.

There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (*i*, *j*) (*i* &lt; *j*) exists such that <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4ac88894f64996c633150129245e8e942fd8c87a.png" style="max-width: 100.0%;max-height: 100.0%;"/> or <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/97e3fd994f862defa2243881d96940d53a455b35.png" style="max-width: 100.0%;max-height: 100.0%;"/>, where <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7be2c0544441f4163b5555e57a9890557fb84ce9.png" style="max-width: 100.0%;max-height: 100.0%;"/> denotes intersection of sets, and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b2ae3cdd5aef4a90a3f4ab78b56dd65351be9b3c.png" style="max-width: 100.0%;max-height: 100.0%;"/> denotes the empty set.

You are to determine whether a valid sequence of operations exists that produces a given final grid.
The first line contains two space-separated integers *n* and *m* (1 ≤ *n*, *m* ≤ 50) — the number of rows and columns of the grid, respectively.

Each of the following *n* lines contains a string of *m* characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
For the first example, the desired setup can be produced by 3 operations, as is shown below.

For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column.
5 8
.#.#..#.
.....#..
.#.#..#.
#.#....#
.....#..
Yes
5 5
..#..
..#..
#####
..#..
..#..
No
5 9
........#
#........
..##.#...
.......#.
....#.#.#
No
1 1
#
Yes
2 1
.
#
Yes
2 5
.####
#..##
No
5 2
##
##
..
##
..
Yes
5 2
#.
##
##
#.
..
No
4 10
###..#..##
...##..#..
.##..#..#.
.........#
No
4 10
..#......#
.....##...
#.........
.#.......#
No
10 15
.......#.......
.....#.........
....#..........
....#..........
.....#.........
.....#.........
#.............#
...#..#........
...............
.............#.
Yes
50 1
.
.
#
.
#
.
#
.
.
#
#
#
.
#
#
#
#
.
.
.
.
.
.
.
.
.
#
#
#
#
.
.
.
.
.
#
.
.
.
#
#
.
.
#
#
.
.
#
#
.
Yes
2 50
...#.##.###...#.#..##....##..........#.#..#.#####.
...#.##.###...#.#..##....##..........#.#..#.#####.
Yes
50 2
..
..
#.
..
.#
..
..
..
.#
..
..
.#
##
..
..
..
.#
..
..
.#
..
..
.#
..
..
.#
..
.#
..
.#
..
.#
.#
.#
..
..
..
.#
.#
.#
..
..
.#
.#
..
..
..
..
..
..
No
1 1
.
Yes
2 3
#.#
###
No
2 3
#.#
##.
No
4 4
###.
##.#
#.##
.###
No
3 3
.##
#.#
##.
No
2 2
##
#.
No
n, m = map(int, input().split())
t = [0]*n

for j in range(n):
    s = input()
    for i in range(len(s)):
        t[j] += int(s[i] == "#")
        t[j] *= 2
        
ans = True
for i in range(n):
    for j in range(i + 1, n):
        if (t[i] & t[j]) != 0:
            if t[i] != t[j]:
                ans = False

if ans:
    print("Yes")
else:
    print("No")