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;
template <class T>
inline void ri(T &i) {
  char c;
  for (c = getchar(); c < '0' || c > '9'; c = getchar())
    ;
  for (i = 0; c >= '0' && c <= '9'; c = getchar())
    i = (i << 3) + (i << 1) + (c - 48);
}
long long n;
char a[1005][1005];
double lg[1005][1005];
long long pre[8][1005][1005];
double sum[8][1005][1005];
int dir[8][2] = {{-1, -1}, {-1, 1}, {1, -1}, {1, 1},
                 {0, -1},  {0, 1},  {1, 0},  {-1, 0}};
bool bounds(long long x) { return (x >= 0 && x < n); }
int main() {
  ri(n);
  int mx = 0;
  double lg2 = log10(2), lg3 = log10(3);
  for (int i = 0; i < n; i++) scanf("%s", a[i]);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      if (a[i][j] == '2')
        lg[i][j] = lg2;
      else if (a[i][j] == '3')
        lg[i][j] = lg3;
      else
        lg[i][j] = 0.000000;
      int dig = a[i][j] - '0';
      mx = max(mx, dig);
    }
  for (int k = 0; k < 8; k++) {
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) {
        if (!bounds(i + dir[k][0]) || !bounds(j + dir[k][1])) {
          int dx = -dir[k][0], dy = -dir[k][1];
          int cnt = 0;
          for (int x = i, y = j; bounds(x) && bounds(y); x += dx, y += dy) {
            if (a[x][y] != '0') {
              cnt++;
              if (x == i && y == j)
                sum[k][x][y] = lg[x][y];
              else
                sum[k][x][y] = sum[k][x - dx][y - dy] + lg[x][y];
            } else {
              cnt = 0;
              sum[k][x][y] = 0.000000;
            }
            pre[k][x][y] = cnt;
          }
        }
      }
  }
  double ans = 0.000000;
  long long ansx, ansy, ansr, ansd;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      if (a[i][j] == '0') continue;
      long long tmp = n + 5;
      for (int k = 0; k < 4; k++) tmp = min(tmp, pre[k][i][j]);
      double tot = lg[i][j];
      for (int k = 0; k < 4; k++) {
        int ix = i + dir[k][0] * (tmp - 1);
        int iy = j + dir[k][1] * (tmp - 1);
        double diff = sum[k][i][j] - (sum[k][ix][iy] + lg[i][j]) + lg[ix][iy];
        tot += diff;
      }
      if (tot > ans) {
        ans = tot;
        ansx = i;
        ansy = j;
        ansr = tmp;
        ansd = 0;
      }
      tmp = n + 5;
      for (int k = 4; k < 8; k++) tmp = min(tmp, pre[k][i][j]);
      tot = lg[i][j];
      for (int k = 4; k < 8; k++) {
        int ix = i + dir[k][0] * (tmp - 1);
        int iy = j + dir[k][1] * (tmp - 1);
        double diff = sum[k][i][j] - (sum[k][ix][iy] + lg[i][j]) + lg[ix][iy];
        tot += diff;
      }
      if (tot > ans) {
        ans = tot;
        ansx = i;
        ansy = j;
        ansr = tmp;
        ansd = 4;
      }
    }
  if (ans == 0.000000) {
    cout << mx << "\n";
    return 0;
  }
  long long total = a[ansx][ansy] - '0';
  for (int k = ansd; k < ansd + 4; k++) {
    for (int i = 1; i < ansr; i++) {
      int ix = ansx + dir[k][0] * i;
      int iy = ansy + dir[k][1] * i;
      int dig = a[ix][iy] - '0';
      total = total * dig;
      total %= 1000000007;
    }
  }
  cout << total << "\n";
  return 0;
}
