Problem: There is a square-shaped grid with N vertical rows and N horizontal columns. We will denote the square at the i-th row from the top and the j-th column from the left as (i,\ j).

Initially, each square is either white or black. The initial color of the grid is given to you as characters a_{ij}, arranged in a square shape. If the square (i,\ j) is white, a_{ij} is `.`. If it is black, a_{ij} is `#`.

You are developing a robot that repaints the grid. It can repeatedly perform the following operation:

* Select two integers i, j (1 ≤ i,\ j ≤ N). Memorize the colors of the squares (i,\ 1), (i,\ 2), ..., (i,\ N) as c_1, c_2, ..., c_N, respectively. Then, repaint the squares (1,\ j), (2,\ j), ..., (N,\ j) with the colors c_1, c_2, ..., c_N, respectively.



Your objective is to turn all the squares black. Determine whether it is possible, and find the minimum necessary number of operations to achieve it if the answer is positive.

Constraints

* 2 ≤ N ≤ 500
* a_{ij} is either `.` or `#`.

Input

The input is given from Standard Input in the following format:


N
a_{11}...a_{1N}
:
a_{N1}...a_{NN}


Output

If it is possible to turn all the squares black, print the minimum necessary number of operations to achieve the objective. If it is impossible, print `-1` instead.

Examples

Input

2
#.
.#


Output

3


Input

2
.
.#


Output

3


Input

2
..
..


Output

-1


Input

2


Output

0


Input

3
.#.

.#.


Output

2


Input

3
...
.#.
...


Output

5

Solution (in PYTHON3):

Solution:
# - どの横一行を揃えるかで場合分け
#   - 上から i 行目を揃える場合、左から i 列目が # になっている行が必要
#     - もともとそのような行があるなら、i 行目を揃えるのに必要な操作回数は
#        N - (初期状態の i 行目の # の数)
#     - そのような行が無い場合、# のある適当な行を i 列目に置くことで作る
#       - 操作回数は N - (初期状態の i 行目の # の数) + 1
# - 横が一行以上揃ったら、縦を揃える
#   - 操作回数は N - (初期状態で揃ってる縦列の数)

N = int(input())
A = [input().replace("#", "1").replace(".", "0") for _ in range(N)]
A_T_bit = [int("".join(a), 2) for a in zip(*A)]

if A_T_bit.count(0) == N:
    print(-1)
    exit()
ans = float("inf")
for row, col in zip(A, A_T_bit):
    if col > 0:
        ans = min(ans, N - row.count("1"))
    else:
        ans = min(ans, N - row.count("1") + 1)
ans += N - A_T_bit.count((1<<N)-1)
print(ans)