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 long long inf = 1e18;
long long n;
long long cost[5], dp[1000 + 10][8][8][8][8], mask[5];
char c[5][1000 + 10];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (long long i = 0; i < 1010; ++i) {
    for (long long m1 = 0; m1 < 8; ++m1) {
      for (long long m2 = 0; m2 < 8; ++m2) {
        for (long long m3 = 0; m3 < 8; ++m3) {
          for (long long m4 = 0; m4 < 8; ++m4) {
            dp[i][m1][m2][m3][m4] = inf;
          }
        }
      }
    }
  }
  cin >> n;
  for (long long i = 1; i <= 4; ++i) cin >> cost[i];
  for (long long i = 1; i < 5; ++i) {
    for (long long j = 1; j <= n; ++j) {
      cin >> c[i][j];
    }
  }
  for (long long i = 1; i < 5; ++i) {
    for (long long j = n + 1; j <= n + 3; ++j) {
      c[i][j] = '.';
    }
  }
  for (long long i = 1; i < 5; ++i) {
    for (long long j = 1; j <= 3; ++j) {
      if (c[i][j] == '.') mask[i] |= (1 << (3 - j));
    }
  }
  dp[3][mask[1]][mask[2]][mask[3]][mask[4]] = 0;
  for (long long i = 3; i < n + 3; ++i) {
    for (long long m1 = 0; m1 < 8; ++m1) {
      for (long long m2 = 0; m2 < 8; ++m2) {
        for (long long m3 = 0; m3 < 8; ++m3) {
          for (long long m4 = 0; m4 < 8; ++m4) {
            if (dp[i][m1][m2][m3][m4] != inf) {
              long long next1, next2, next3, next4;
              next1 = (1 << 2) | m1;
              next2 = (1 << 2) | m2;
              next3 = (1 << 2) | m3;
              next4 = (1 << 2) | m4;
              dp[i][next1][m2][m3][m4] = min(dp[i][next1][m2][m3][m4],
                                             dp[i][m1][m2][m3][m4] + cost[1]);
              dp[i][m1][next2][m3][m4] = min(dp[i][m1][next2][m3][m4],
                                             dp[i][m1][m2][m3][m4] + cost[1]);
              dp[i][m1][m2][next3][m4] = min(dp[i][m1][m2][next3][m4],
                                             dp[i][m1][m2][m3][m4] + cost[1]);
              dp[i][m1][m2][m3][next4] = min(dp[i][m1][m2][m3][next4],
                                             dp[i][m1][m2][m3][m4] + cost[1]);
              next1 |= 2;
              next2 |= 2;
              next3 |= 2;
              next4 |= 2;
              dp[i][next1][next2][m3][m4] = min(
                  dp[i][next1][next2][m3][m4], dp[i][m1][m2][m3][m4] + cost[2]);
              dp[i][m1][next2][next3][m4] = min(
                  dp[i][m1][next2][next3][m4], dp[i][m1][m2][m3][m4] + cost[2]);
              dp[i][m1][m2][next3][next4] = min(
                  dp[i][m1][m2][next3][next4], dp[i][m1][m2][m3][m4] + cost[2]);
              next1 |= 1;
              next2 |= 1;
              next3 |= 1;
              next4 |= 1;
              dp[i][next1][next2][next3][m4] =
                  min(dp[i][next1][next2][next3][m4],
                      dp[i][m1][m2][m3][m4] + cost[3]);
              dp[i][m1][next2][next3][next4] =
                  min(dp[i][m1][next2][next3][next4],
                      dp[i][m1][m2][m3][m4] + cost[3]);
              if ((m1 & 4) && (m2 & 4) && (m3 & 4) && (m4 & 4)) {
                next1 = ((m1 << 1) & 7);
                if (c[1][i + 1] == '.') next1 |= 1;
                next2 = ((m2 << 1) & 7);
                if (c[2][i + 1] == '.') next2 |= 1;
                next3 = ((m3 << 1) & 7);
                if (c[3][i + 1] == '.') next3 |= 1;
                next4 = ((m4 << 1) & 7);
                if (c[4][i + 1] == '.') next4 |= 1;
                dp[i + 1][next1][next2][next3][next4] =
                    min(dp[i + 1][next1][next2][next3][next4],
                        dp[i][m1][m2][m3][m4]);
              }
              dp[i + 1][7][7][7][7] =
                  min(dp[i + 1][7][7][7][7], dp[i][m1][m2][m3][m4] + cost[4]);
            }
          }
        }
      }
    }
  }
  cout << dp[n + 3][7][7][7][7];
  return 0;
}
