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 inf = 1e9 + 3;
int n, n1, vg;
int a[1007];
int dp[1007][(1 << 12)];
char c[4][1007];
vector<pair<int, int> > g;
int b[4][4];
int smm;
void prov() {
  bool f = true;
  for (int i = 0; i < 4; ++i) {
    if (b[i][0] == 1) f = false;
  }
  if (!f) return;
  int mask = 0;
  int cost = smm;
  for (int j = 0; j < 4; ++j) {
    for (int i = 0; i < 4; ++i) {
      if (b[i][j] != 1) continue;
      mask |= (1 << (j * 4 + i));
    }
  }
  mask = (mask >> 4);
  g.push_back(make_pair(mask, cost));
}
void rec1(int num) {
  if (num == 4) {
    prov();
    return;
  }
  int z = min(4, min(4 - (vg - n1), 4 - num));
  int sm1 = 0;
  for (int k = 1; k <= z; ++k) {
    sm1 = 0;
    for (int i = num; i < num + k; ++i) {
      for (int j = 0; j < k; ++j) {
        b[i][j]--;
        sm1 += (b[i][j] == 0);
      }
    }
    if (sm1 == 0) {
      for (int i = num; i < num + k; ++i) {
        for (int j = 0; j < k; ++j) {
          b[i][j]++;
        }
      }
      continue;
    }
    smm += a[k];
    rec1(num + 1);
    smm -= a[k];
    for (int i = num; i < num + k; ++i) {
      for (int j = 0; j < k; ++j) {
        b[i][j]++;
      }
    }
  }
  rec1(num + 1);
}
void rec(int mask) {
  for (int j = 0; j < 4; ++j) {
    for (int i = 0; i < 4; ++i) {
      b[i][j] = ((mask & (1 << (j * 4 + i))) > 0);
    }
  }
  rec1(0);
}
int d[4][4];
void write_msk(int mask) {
  for (int j = 0; j < 4; ++j) {
    for (int i = 0; i < 4; ++i) {
      d[i][j] = ((mask & (1 << (j * 4 + i))) > 0);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= 4; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < 4; ++i) {
    for (int j = 1; j <= n; ++j) {
      cin >> c[i][j];
    }
  }
  int kl = 0;
  int sm = 0;
  for (int j = 1; j <= 3; ++j) {
    for (int i = 0; i < 4; ++i) {
      if (c[i][j] == '*') {
        sm |= (1 << (kl));
      }
      kl++;
    }
  }
  n1 = n;
  n += 3;
  for (int i = 0; i <= n; ++i) {
    for (int j = 0; j < (1 << 12); ++j) {
      dp[i][j] = inf;
    }
  }
  dp[3][sm] = 0;
  for (int i = 4; i <= n; ++i) {
    int sm1 = 0;
    vg = i;
    for (int j = 0; j < 4; ++j) {
      if (c[j][i] == '*') {
        sm1 |= (1 << (12 + j));
      }
    }
    for (int j = 0; j < (1 << 12); ++j) {
      if (dp[i - 1][j] == inf) continue;
      int sm = j + sm1;
      g.clear();
      rec(sm);
      for (int k = 0; k < g.size(); ++k) {
        dp[i][g[k].first] = min(dp[i][g[k].first], dp[i - 1][j] + g[k].second);
      }
    }
  }
  cout << dp[n][0];
}
