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.
#include <bits/stdc++.h>
using namespace std;
const int MAXSIZE = 10000020;
const int INF = 0x3f3f3f3f;
int bufpos;
char buf[MAXSIZE];
void init() {
  buf[fread(buf, 1, MAXSIZE, stdin)] = '\0';
  bufpos = 0;
}
int readint() {
  int val = 0;
  for (; !isdigit(buf[bufpos]); bufpos++)
    ;
  for (; isdigit(buf[bufpos]); bufpos++) val = val * 10 + buf[bufpos] - '0';
  return val;
}
char readchar() {
  for (; isspace(buf[bufpos]); bufpos++)
    ;
  return buf[bufpos++];
}
int readstr(char* s) {
  int cur = 0;
  for (; isspace(buf[bufpos]); bufpos++)
    ;
  for (; !isspace(buf[bufpos]); bufpos++) s[cur++] = buf[bufpos];
  s[cur] = '\0';
  return cur;
}
int dp[1006][5][627], c[6], tmp[6], bk[6];
bool is[1006][5];
int main() {
  init();
  int n = readint();
  for (int i = 1; i <= 4; i++) c[i] = readint();
  for (int i = 0; i < 4; i++)
    for (int j = 1; j <= n; j++) is[j][i] = readchar() == '*';
  memset(dp, 0x3f, sizeof(dp));
  if (is[1][0])
    dp[1][0][1] = 0;
  else
    dp[1][0][0] = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 4; j++) {
      for (int k = 0; k < 625; k++) {
        int qwq = k;
        for (int o = 0; o < 4; o++) bk[o] = qwq % 5, qwq /= 5;
        for (int o = 0; o <= j + 1; o++) {
          for (int t = 0; t < 4; t++) tmp[t] = bk[t];
          for (int t = j - o + 1; t <= j; t++)
            if (tmp[t] <= o) tmp[t] = 0;
          if (j == 3) {
            bool flag = false;
            for (int t = 0; t < 4; t++) {
              if (tmp[t]) tmp[t]++;
              if (tmp[t] >= 5) {
                flag = true;
                break;
              }
            }
            if (flag) continue;
            if (!tmp[0] && is[i + 1][0]) tmp[0] = 1;
            int now = 0;
            for (int t = 3; t >= 0; t--) now = now * 5 + tmp[t];
            dp[i + 1][0][now] = min(dp[i + 1][0][now], dp[i][j][k] + c[o]);
          } else {
            if (!tmp[j + 1] && is[i][j + 1]) tmp[j + 1] = 1;
            int now = 0;
            for (int t = 3; t >= 0; t--) now = now * 5 + tmp[t];
            dp[i][j + 1][now] = min(dp[i][j + 1][now], dp[i][j][k] + c[o]);
          }
        }
      }
    }
  printf("%d", dp[n + 1][0][0]);
}
