Vanya plays a game of balloons on the field of size n × n, where each cell contains a balloon with one of the values 0, 1, 2 or 3. The goal is to destroy a cross, such that the product of all values of balloons in the cross is maximum possible. There are two types of crosses: normal and rotated. For example:
    
    
      
    **o**  
    **o**  
    ooooo  
    **o**  
    **o**  
    

or
    
    
      
    o***o  
    *o*o*  
    **o**  
    *o*o*  
    o***o  
    

Formally, the cross is given by three integers r, c and d, such that d ≤ r, c ≤ n - d + 1. The normal cross consists of balloons located in cells (x, y) (where x stay for the number of the row and y for the number of the column), such that |x - r|·|y - c| = 0 and |x - r| + |y - c| < d. Rotated cross consists of balloons located in cells (x, y), such that |x - r| = |y - c| and |x - r| < d.

Vanya wants to know the maximum possible product of the values of balls forming one cross. As this value can be large, output it modulo 109 + 7.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 1000) — the number of rows and columns in the table with balloons.

The each of the following n lines contains n characters '0', '1', '2' or '3' — the description of the values in balloons.

Output

Print the maximum possible product modulo 109 + 7. Note, that you are not asked to maximize the remainder modulo 109 + 7, but to find the maximum value and print it this modulo.

Examples

Input

4
1233
0213
2020
0303


Output

108


Input

5
00300
00300
33333
00300
00300


Output

19683


Input

5
00003
02030
00300
03020
30000


Output

108


Input

5
21312
10003
10002
10003
23231


Output

3


Input

5
12131
12111
12112
21311
21212


Output

24

Note

In the first sample, the maximum product is achieved for a rotated cross with a center in the cell (3, 3) and radius 1: 2·2·3·3·3 = 108.
#include <bits/stdc++.h>
using namespace std;
double sum[1010][1010][8];
int num[1010][1010][8];
double val2[1010][1010];
char val[1010][1010];
int maxn[1010][1010][2];
int dir[8][2] = {{-1, -1}, {-1, 0}, {-1, 1}, {0, 1},
                 {1, 1},   {1, 0},  {1, -1}, {0, -1}};
int main() {
  int n;
  scanf("%d", &n);
  memset(val2, 0, sizeof(val2));
  for (int i = 1; i <= n; i++) scanf("%s", val[i] + 1);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (val[i][j] == '2')
        val2[i][j] = log(2);
      else if (val[i][j] == '3')
        val2[i][j] = log(3);
      else
        val2[i][j] = 0;
    }
  }
  int sx, sy, x, y;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      for (int k = 0; k < 4; k++) {
        if (k == 0)
          sx = i, sy = j;
        else if (k == 1)
          sx = i, sy = n + 1 - j;
        else if (k == 2)
          sx = n + 1 - i, sy = n + 1 - j;
        else
          sx = n + 1 - i, sy = j;
        for (int l = k * 2; l < k * 2 + 2; l++) {
          x = sx + dir[l][0], y = sy + dir[l][1];
          if (val[sx][sy] == '0')
            num[sx][sy][l] = 0;
          else
            num[sx][sy][l] = num[x][y][l] + 1;
          sum[sx][sy][l] = sum[x][y][l] + val2[sx][sy];
        }
      }
    }
  }
  double ans, maxns = 0;
  int posx = 1, posy = 1, sign = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (val[i][j] == '0') continue;
      maxn[i][j][0] =
          min(num[i][j][0], min(num[i][j][2], min(num[i][j][4], num[i][j][6])));
      x = maxn[i][j][0];
      ans = sum[i][j][0] - sum[i - x][j - x][0] + sum[i][j][2] -
            sum[i - x][j + x][2] + sum[i][j][4] - sum[i + x][j + x][4] +
            sum[i][j][6] - sum[i + x][j - x][6];
      ans -= 3 * val2[i][j];
      if (ans > maxns) {
        maxns = ans;
        posx = i;
        posy = j;
        sign = 0;
      }
      maxn[i][j][1] =
          min(num[i][j][1], min(num[i][j][3], min(num[i][j][5], num[i][j][7])));
      x = maxn[i][j][1];
      ans = sum[i][j][1] - sum[i - x][j][1] + sum[i][j][3] - sum[i][j + x][3] +
            sum[i][j][5] - sum[i + x][j][5] + sum[i][j][7] - sum[i][j - x][7];
      ans -= 3 * val2[i][j];
      if (ans > maxns) {
        maxns = ans;
        posx = i;
        posy = j;
        sign = 1;
      }
    }
  }
  long long fin = val[posx][posy] - '0';
  for (int i = 1; i < maxn[posx][posy][sign]; i++) {
    for (int j = 0; j < 4; j++) {
      x = posx + i * dir[j * 2 + sign][0], y = posy + i * dir[j * 2 + sign][1];
      fin *= val[x][y] - '0';
      fin %= 1000000007;
    }
  }
  printf("%I64d\n", fin);
  return 0;
}
