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;
const int MAXN = 1E3 + 10;
const int MOD = 1E9 + 7;
int n;
char s[MAXN][MAXN];
double p[4][MAXN][MAXN];
int a[8][MAXN][MAXN];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%s", s[i] + 1);
  double val[4] = {0., 0., log(2.), log(3.)};
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j) {
      s[i][j] -= '0';
      p[0][i][j] = p[0][i][j - 1] + val[s[i][j]];
      p[1][i][j] = p[1][i - 1][j - 1] + val[s[i][j]];
      p[2][i][j] = p[2][i - 1][j] + val[s[i][j]];
      p[3][i][j] = p[3][i - 1][j + 1] + val[s[i][j]];
    }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      if (s[i][j]) {
        a[0][i][j] = a[0][i][j - 1] + 1;
        a[1][i][j] = a[1][i - 1][j - 1] + 1;
        a[2][i][j] = a[2][i - 1][j] + 1;
        a[3][i][j] = a[3][i - 1][j + 1] + 1;
      } else
        a[0][i][j] = a[1][i][j] = a[2][i][j] = a[3][i][j] = 0;
  for (int i = n; i >= 1; --i)
    for (int j = n; j >= 1; --j)
      if (s[i][j]) {
        a[4][i][j] = a[4][i][j + 1] + 1;
        a[5][i][j] = a[5][i + 1][j + 1] + 1;
        a[6][i][j] = a[6][i + 1][j] + 1;
        a[7][i][j] = a[7][i + 1][j - 1] + 1;
      } else
        a[4][i][j] = a[5][i][j] = a[6][i][j] = a[7][i][j] = 0;
  double ans = -1.0;
  int x = 0, y = 0, z = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j) {
      if (!s[i][j]) continue;
      int l = min(min(a[0][i][j], a[2][i][j]), min(a[4][i][j], a[6][i][j]));
      double res = p[0][i][j + l - 1] - p[0][i][j - l] + p[2][i + l - 1][j] -
                   p[2][i - l][j] - val[s[i][j]];
      if (ans < res) ans = res, x = i, y = j, z = l;
      l = min(min(a[1][i][j], a[3][i][j]), min(a[5][i][j], a[7][i][j]));
      res = p[1][i + l - 1][j + l - 1] - p[1][i - l][j - l] +
            p[3][i + l - 1][j - l + 1] - p[3][i - l][j + l] - val[s[i][j]];
      if (ans < res) ans = res, x = i, y = j, z = -l;
    }
  if (ans < 0.)
    puts("0");
  else if (z > 0) {
    unsigned ans = s[x][y];
    for (int i = -z + 1; i < z; ++i)
      if (i) {
        ans = ans * s[x][y + i] % MOD;
        ans = ans * s[x + i][y] % MOD;
      }
    printf("%u\n", ans);
  } else {
    unsigned ans = s[x][y];
    for (int i = -z - 1; i > z; --i)
      if (i) {
        ans = ans * s[x + i][y + i] % MOD;
        ans = ans * s[x + i][y - i] % MOD;
      }
    printf("%u\n", ans);
  }
  return 0;
}
