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;
char s[1000 + 5][1000 + 5];
int a[1000 + 5][1000 + 5];
int a2, a3;
bool mark;
int pre[4][1000 + 5][1000 + 5];
int cnt2[2][1000 + 5][1000 + 5], cnt3[2][1000 + 5][1000 + 5];
void solve(int n) {
  for (int i = 1; i <= n; i++) pre[0][i][0] = 0;
  for (int i = 1; i <= n; i++) pre[1][i][n + 1] = n + 1;
  for (int i = 1; i <= n; i++) pre[2][0][i] = 0;
  for (int i = 1; i <= n; i++) pre[3][n + 1][i] = n + 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      pre[0][i][j] = a[i][j] == 0 ? j : pre[0][i][j - 1];
  for (int i = 1; i <= n; i++)
    for (int j = n; j >= 1; j--)
      pre[1][i][j] = a[i][j] == 0 ? j : pre[1][i][j + 1];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      pre[2][i][j] = a[i][j] == 0 ? i : pre[2][i - 1][j];
  for (int i = n; i >= 1; i--)
    for (int j = 1; j <= n; j++)
      pre[3][i][j] = a[i][j] == 0 ? i : pre[3][i + 1][j];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      cnt2[0][i][j] = cnt2[0][i][j - 1] + (a[i][j] == 2);
      cnt2[1][i][j] = cnt2[1][i - 1][j] + (a[i][j] == 2);
      cnt3[0][i][j] = cnt3[0][i][j - 1] + (a[i][j] == 3);
      cnt3[1][i][j] = cnt3[1][i - 1][j] + (a[i][j] == 3);
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (a[i][j]) {
        int len = 0x3f3f3f3f;
        len = min(len, j - pre[0][i][j]);
        len = min(len, pre[1][i][j] - j);
        len = min(len, i - pre[2][i][j]);
        len = min(len, pre[3][i][j] - i);
        int l = j - len + 1, r = j + len - 1, u = i - len + 1, d = i + len - 1;
        int c2 = cnt2[0][i][r] - cnt2[0][i][l - 1] + cnt2[1][d][j] -
                 cnt2[1][u - 1][j] - (a[i][j] == 2);
        int c3 = cnt3[0][i][r] - cnt3[0][i][l - 1] + cnt3[1][d][j] -
                 cnt3[1][u - 1][j] - (a[i][j] == 3);
        int d2 = c2 - a2, d3 = c3 - a3;
        if (d2 + log2(3) * d3 >= 0) a2 = c2, a3 = c3, mark = 1;
      }
}
long long qp(long long a, int k) {
  long long res = 1;
  while (k) {
    if (k & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
    k >>= 1;
  }
  return res;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%s", s[i] + 1);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) a[i][j] = s[i][j] - '0';
  solve(n);
  int h = (n + 1) / 2;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) a[i][j] = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if ((i + j) % 2 == 0) a[(i + j) / 2][h + (j - i) / 2] = s[i][j] - '0';
  solve(n);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) a[i][j] = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if ((i + j) % 2 == 1) a[(i + j) / 2][h + (j - i + 1) / 2] = s[i][j] - '0';
  solve(n);
  if (mark)
    printf("%I64d\n", qp(2, a2) * qp(3, a3) % 1000000007);
  else
    puts("0");
  return 0;
}
