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 double lg2 = log(2.0);
const double lg3 = log(3.0);
const int dir[][2] = {{-1, -1}, {-1, 1}, {1, -1}, {1, 1},
                      {0, 1},   {0, -1}, {1, 0},  {-1, 0}};
const long long mod = 1e9 + 7;
const int mxn = 1000 + 10;
int a[mxn][mxn];
int zero[8][mxn][mxn];
double lg[mxn][mxn];
double sum[8][mxn][mxn];
int n;
inline bool inboard(int x, int y) {
  return x >= 1 && x <= n && y >= 1 && y <= n;
}
long long solve() {
  for (int k = 0; k < 8; k++) {
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        if (!inboard(i - dir[k][0], j - dir[k][1])) {
          double s = 0;
          int zcnt = 0;
          for (int x = i, y = j; inboard(x, y);
               x += dir[k][0], y += dir[k][1]) {
            if (a[x][y] == 0) {
              s = 0;
              zcnt = 0;
            } else {
              zcnt++;
              s += lg[x][y];
            }
            sum[k][x][y] = s;
            zero[k][x][y] = zcnt;
          }
        }
  }
  double ans = -1;
  int ansx = 1, ansy = 1, ansd = 1, ansk = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++)
      if (a[i][j])
        for (int k = 0; k <= 4; k += 4) {
          int td = min(min(zero[k][i][j], zero[k + 1][i][j]),
                       min(zero[k + 2][i][j], zero[k + 3][i][j]));
          double snow = lg[i][j];
          for (int dr = 0; dr < 4; dr++)
            snow += sum[k + dr][i + dir[k + dr][0] * (td - 1)]
                       [j + dir[k + dr][1] * (td - 1)] -
                    sum[k + dr][i][j];
          if (snow > ans) {
            ans = snow;
            ansx = i;
            ansy = j;
            ansd = td;
            ansk = k;
          }
        }
  }
  long long ret = a[ansx][ansy];
  for (int k = ansk; k < ansk + 4; k++) {
    for (int i = 1; i < ansd; i++) {
      ret = ret * (a[ansx + dir[k][0] * i][ansy + dir[k][1] * i]) % mod;
    }
  }
  return ret;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      char t;
      cin >> t;
      a[i][j] = t - '0';
      if (t == '2')
        lg[i][j] = lg2;
      else if (t == '3')
        lg[i][j] = lg3;
    }
  }
  cout << solve() << endl;
  return 0;
}
