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[10050];
int a[1200][1200];
int z[8][1200][1200];
pair<int, int> tu[1200][1200];
pair<int, int> tl[1200][1200];
pair<int, int> tdl[1200][1200];
pair<int, int> tul[1200][1200];
pair<int, int> sp[2200200];
const double ln2 = log(double(2));
const double ln3 = log(double(3));
const int mod = int(1e9) + 7;
int main() {
  int n;
  gets(s);
  sscanf(s, "%d", &n);
  for (int i = 0; i < n; i++) {
    gets(s);
    for (int j = 0; j < n; j++) a[i + 1][j + 1] = s[j] - '0';
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (a[i][j]) {
        z[0][i][j] = z[0][i - 1][j] + 1;
        z[1][i][j] = z[1][i][j - 1] + 1;
        z[4][i][j] = z[4][i - 1][j - 1] + 1;
        z[5][i][j] = z[5][i - 1][j + 1] + 1;
        tu[i][j] = tu[i - 1][j];
        tl[i][j] = tl[i][j - 1];
        tul[i][j] = tul[i - 1][j - 1];
        if (a[i][j] == 2) {
          tu[i][j].first++;
          tl[i][j].first++;
          tul[i][j].first++;
        } else if (a[i][j] == 3) {
          tu[i][j].second++;
          tl[i][j].second++;
          tul[i][j].second++;
        }
      }
  for (int i = n; i > 0; i--)
    for (int j = n; j > 0; j--)
      if (a[i][j]) {
        z[2][i][j] = z[2][i + 1][j] + 1;
        z[3][i][j] = z[3][i][j + 1] + 1;
        z[6][i][j] = z[6][i + 1][j + 1] + 1;
        z[7][i][j] = z[7][i + 1][j - 1] + 1;
        tdl[i][j] = tdl[i + 1][j - 1];
        if (a[i][j] == 2) {
          tdl[i][j].first++;
        } else if (a[i][j] == 3) {
          tdl[i][j].second++;
        }
      }
  int pos = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (a[i][j]) {
        int d = 1e9;
        for (int k = 0; k < 4; k++)
          if (z[k][i][j] < d) d = z[k][i][j];
        d--;
        int x = tl[i][j + d].first - tl[i][j - d - 1].first +
                tu[i + d][j].first - tu[i - d - 1][j].first;
        int y = tl[i][j + d].second - tl[i][j - d - 1].second +
                tu[i + d][j].second - tu[i - d - 1][j].second;
        if (a[i][j] == 2)
          x--;
        else if (a[i][j] == 3)
          y--;
        sp[pos++] = make_pair(x, y);
        d = 1e9;
        for (int k = 4; k < 8; k++)
          if (z[k][i][j] < d) d = z[k][i][j];
        d--;
        x = tdl[i - d][j + d].first - tdl[i + d + 1][j - d - 1].first +
            tul[i + d][j + d].first - tul[i - d - 1][j - d - 1].first;
        y = tdl[i - d][j + d].second - tdl[i + d + 1][j - d - 1].second +
            tul[i + d][j + d].second - tul[i - d - 1][j - d - 1].second;
        if (a[i][j] == 2)
          x--;
        else if (a[i][j] == 3)
          y--;
        sp[pos++] = make_pair(x, y);
      }
  if (pos == 0) {
    cout << 0 << endl;
    return 0;
  }
  sort(sp, sp + pos);
  int k = 0;
  for (int i = 0; i < pos; i++) {
    while (k > 0 && sp[k - 1].second <= sp[i].second) k--;
    sp[k] = sp[i];
    k++;
  }
  int ra = sp[0].first;
  int rb = sp[0].second;
  for (int i = 1; i < k; i++)
    if ((sp[i].first - ra) * ln2 > (rb - sp[i].second) * ln3) {
      ra = sp[i].first;
      rb = sp[i].second;
    }
  long long res = 1;
  while (ra--) res = (res * 2) % mod;
  while (rb--) res = (res * 3) % mod;
  cout << res << endl;
  return 0;
}
