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;
int n, r, a[1050][1050], len[1050][1050][10];
double mx = -1, res, sum[1050][1050][10];
char s[1050][1050];
long long ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%s", s[i] + 1);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) a[i][j] = s[i][j] - '0';
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (a[i][j]) {
        sum[i][j][0] = sum[i - 1][j - 1][0] + log(a[i][j]);
        sum[i][j][1] = sum[i - 1][j][1] + log(a[i][j]);
        sum[i][j][2] = sum[i - 1][j + 1][2] + log(a[i][j]);
        sum[i][j][3] = sum[i][j - 1][3] + log(a[i][j]);
        len[i][j][0] = len[i - 1][j - 1][0] + 1;
        len[i][j][1] = len[i - 1][j][1] + 1;
        len[i][j][2] = len[i - 1][j + 1][2] + 1;
        len[i][j][3] = len[i][j - 1][3] + 1;
      }
  for (int i = n; i >= 1; i--)
    for (int j = n; j >= 1; j--)
      if (a[i][j]) {
        len[i][j][5] = len[i][j + 1][5] + 1;
        len[i][j][6] = len[i + 1][j - 1][6] + 1;
        len[i][j][7] = len[i + 1][j][7] + 1;
        len[i][j][8] = len[i + 1][j + 1][8] + 1;
      }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (a[i][j]) {
        r = min(min(len[i][j][0], len[i][j][2]),
                min(len[i][j][6], len[i][j][8])) -
            1;
        res = sum[i + r][j + r][0] - sum[i - r - 1][j - r - 1][0] +
              sum[i + r][j - r][2] - sum[i - r - 1][j + r + 1][2] -
              log(a[i][j]);
        if (mx < res) {
          mx = res;
          ans = a[i][j];
          for (int k = -r; k <= r; k++)
            if (k)
              ans = ans * a[i + k][j + k] % 1000000007 * a[i - k][j + k] %
                    1000000007;
        }
        r = min(min(len[i][j][1], len[i][j][3]),
                min(len[i][j][5], len[i][j][7])) -
            1;
        res = sum[i + r][j][1] - sum[i - r - 1][j][1] + sum[i][j + r][3] -
              sum[i][j - r - 1][3] - log(a[i][j]);
        if (mx < res) {
          mx = res;
          ans = a[i][j];
          for (int k = -r; k <= r; k++)
            if (k)
              ans = ans * a[i][j + k] % 1000000007 * a[i + k][j] % 1000000007;
        }
      }
  printf("%lld\n", ans);
  return 0;
}
