903_F. Clear The Matrix
You are given a matrix f with 4 rows and n columns. Each element of the matrix is either an asterisk (*) or a dot (.).

You may perform the following operation arbitrary number of times: choose a square submatrix of f with size k × k (where 1 ≤ k ≤ 4) and replace each element of the chosen submatrix with a dot. Choosing a submatrix of size k × k costs ak coins.

What is the minimum number of coins you have to pay to replace all asterisks with dots?

Input

The first line contains one integer n (4 ≤ n ≤ 1000) — the number of columns in f.

The second line contains 4 integers a1, a2, a3, a4 (1 ≤ ai ≤ 1000) — the cost to replace the square submatrix of size 1 × 1, 2 × 2, 3 × 3 or 4 × 4, respectively.

Then four lines follow, each containing n characters and denoting a row of matrix f. Each character is either a dot or an asterisk.

Output

Print one integer — the minimum number of coins to replace all asterisks with dots.

Examples

Input

4
1 10 8 20
***.
***.
***.
...*


Output

9


Input

7
2 1 8 2
.***...
.***..*
.***...
....*..


Output

3


Input

4
10 10 1 10
***.
*..*
*..*
.***


Output

2

Note

In the first example you can spend 8 coins to replace the submatrix 3 × 3 in the top-left corner, and 1 coin to replace the 1 × 1 submatrix in the bottom-right corner.

In the second example the best option is to replace the 4 × 4 submatrix containing columns 2 – 5, and the 2 × 2 submatrix consisting of rows 2 – 3 and columns 6 – 7.

In the third example you can select submatrix 3 × 3 in the top-left corner and then submatrix 3 × 3 consisting of rows 2 – 4 and columns 2 – 4.
```python
import sys

n = int(sys.stdin.buffer.readline().decode('utf-8'))
cost = [0] + \
    list(map(int, sys.stdin.buffer.readline().decode('utf-8').split()))
a = [line.decode('utf-8').rstrip() for line in sys.stdin.buffer]

mask = [0, 1, 51, 1911]
inf, bs_size, full_bit = 10**9, 1 << 12, (1 << 12) - 1
dp = [[inf]*bs_size for _ in range(4*n+1)]
dp[0][0] = 0

for i in range(4*n):
    y, x = i & 3, i >> 2
    is_dot = 1 if a[y][x] == '.' else 0

    for bitset in range(bs_size):
        if y == 0:
            '''
              01234    01234
            0 s****    .t..*
            1 ***** -> ....*
            2 ***** -> ....*
            3 *****    ....*
            '''
            if dp[i+4][full_bit] > dp[i][bitset] + cost[4]:
                dp[i+4][full_bit] = dp[i][bitset] + cost[4]

        if (is_dot | bitset & 1) and\
                dp[i+1][bitset >> 1] > dp[i][bitset]:
            dp[i+1][bitset >> 1] = dp[i][bitset]

        for k in range(1, min(4-y, 3)+1):
            if dp[i][bitset | mask[k]] > dp[i][bitset] + cost[k]:
                dp[i][bitset | mask[k]] = dp[i][bitset] + cost[k]

print(min(dp[4*n]))

```