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 + 4;
const int BASE = 1e9 + 7;
char a[MAXN][MAXN];
double logs[MAXN][MAXN], sum[9][MAXN][MAXN], log_2 = log(2.0), log_3 = log(3.0),
                                             ans = -1;
int n, pre[9][MAXN][MAXN], ansx, ansy, anssize, ansdir;
int dx[] = {-1, -1, 1, 1, -1, 0, 1, 0};
int dy[] = {-1, 1, 1, -1, 0, 1, 0, -1};
bool Inside(int x, int y) { return (1 <= x && x <= n && 1 <= y && y <= n); }
int main() {
  scanf("%d\n", &n);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      char c = getchar();
      a[i][j] = c;
      if (a[i][j] == '0')
        logs[i][j] = 0;
      else if (a[i][j] == '1')
        logs[i][j] = 0;
      else if (a[i][j] == '2')
        logs[i][j] = log_2;
      else
        logs[i][j] = log_3;
    }
    scanf("\n");
  }
  for (int dir = 0; dir <= 7; ++dir) {
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= n; ++j)
        if (!Inside(i - dx[dir], j - dy[dir])) {
          int d1 = dx[dir], d2 = dy[dir];
          for (int x = i, y = j; Inside(x, y); x += d1, y += d2) {
            if (a[x][y] != '0') {
              if (x == i && y == j) {
                sum[dir][x][y] = logs[x][y];
                pre[dir][x][y] = 1;
              } else {
                sum[dir][x][y] = sum[dir][x - d1][y - d2] + logs[x][y];
                pre[dir][x][y] = pre[dir][x - d1][y - d2] + 1;
              }
            } else {
              sum[dir][x][y] = 0;
              pre[dir][x][y] = 0;
            }
          }
        }
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      if (a[i][j] != '0') {
        int temp = pre[0][i][j];
        for (int k = 1; k <= 3; ++k) temp = min(temp, pre[k][i][j]);
        double val = logs[i][j];
        for (int k = 0; k <= 3; ++k)
          val += sum[k][i + dx[k] * (temp - 1)][j + dy[k] * (temp - 1)] -
                 sum[k][i][j];
        if (val > ans) {
          ans = val;
          ansx = i;
          ansy = j;
          anssize = temp;
          ansdir = 0;
        }
        temp = pre[4][i][j];
        for (int k = 5; k <= 7; ++k) temp = min(temp, pre[k][i][j]);
        val = logs[i][j];
        for (int k = 4; k <= 7; ++k)
          val += sum[k][i + dx[k] * (temp - 1)][j + dy[k] * (temp - 1)] -
                 sum[k][i][j];
        if (val > ans) {
          ans = val;
          ansx = i;
          ansy = j;
          anssize = temp;
          ansdir = 4;
        }
      }
  if (ans == -1)
    putchar('0');
  else {
    int total = a[ansx][ansy] - '0';
    for (int dir = ansdir; dir <= ansdir + 3; ++dir) {
      for (int i = 2; i <= anssize; ++i) {
        total =
            (1ll * total *
             (a[ansx + dx[dir] * (i - 1)][ansy + dy[dir] * (i - 1)] - '0')) %
            BASE;
      }
    }
    printf("%d", total);
  }
}
