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 a[8] = {0, -1, -1, -1, 0, 1, 1, 1},
          b[8] = {-1, -1, 0, 1, 1, 1, 0, -1}, MOD = 1e9 + 7, N = 1e3 + 99;
int n, g[N][N], d[8][N][N];
double lg[8][N][N];
char ch;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) cin >> ch, g[i][j] = ch - '0';
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      for (int k = 0; k < 4; k++)
        if (!g[i][j])
          d[k][i][j] = 0, lg[k][i][j] = -1;
        else
          d[k][i][j] = d[k][i + a[k]][j + b[k]] + 1,
          lg[k][i][j] = lg[k][i + a[k]][j + b[k]] + log(g[i][j]);
  for (int i = n; i > 0; i--)
    for (int j = n; j > 0; j--)
      for (int k = 4; k < 8; k++)
        if (!g[i][j])
          d[k][i][j] = 0, lg[k][i][j] = -1;
        else
          d[k][i][j] = d[k][i + a[k]][j + b[k]] + 1,
          lg[k][i][j] = lg[k][i + a[k]][j + b[k]] + log(g[i][j]);
  double maxlg = -1;
  int x, y, rr = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      int r;
      r = min(min(d[0][i][j], d[2][i][j]), min(d[4][i][j], d[6][i][j]));
      double tmp = 0;
      for (int k = 0; k < 8; k += 2)
        tmp += lg[k][i][j] - lg[k][i + a[k] * r][j + b[k] * r];
      if (g[i][j]) tmp -= log(g[i][j]) * 3;
      if (maxlg < tmp) maxlg = tmp, x = i, y = j, rr = r;
      r = min(min(d[1][i][j], d[3][i][j]), min(d[5][i][j], d[7][i][j]));
      tmp = 0;
      for (int k = 1; k < 8; k += 2)
        tmp += lg[k][i][j] - lg[k][i + a[k] * r][j + b[k] * r];
      if (g[i][j]) tmp -= log(g[i][j]) * 3;
      if (maxlg < tmp) maxlg = tmp, x = i, y = j, rr = -r;
    }
  long long ans = 0;
  if (rr > 0) {
    ans = g[x][y];
    for (int k = 0; k < 8; k += 2)
      for (int i = 1; i < rr; i++)
        ans = (ans * g[x + a[k] * i][y + b[k] * i]) % MOD;
  } else if (rr < 0) {
    ans = g[x][y];
    for (int k = 1; k < 8; k += 2)
      for (int i = 1; i < -rr; i++)
        ans = (ans * g[x + a[k] * i][y + b[k] * i]) % MOD;
  }
  cout << ans;
  return 0;
}
