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;
vector<pair<int, int> > v;
int i, j, k, n, m, x, y, z;
char a[1005][1005];
int H[2005][2005], l;
int A[1005][1005][3][9];
int yon[8][2] = {0, -1, -1, -1, -1, 0, -1, 1, 0, 1, 1, 1, 1, 0, 1, -1};
bool cmp(pair<int, int> x, pair<int, int> y) {
  int t1 = min(x.first, y.first);
  x.first -= t1;
  y.first -= t1;
  int t2 = min(x.second, y.second);
  x.second -= t2;
  y.second -= t2;
  if (!y.first and !y.second) return 0;
  if (!x.first and !x.second) return 1;
  if (x.first and y.second) return H[y.second][x.first];
  return !H[x.second][y.first];
}
int main() {
  cin >> n;
  int p = 0;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) {
      scanf(" %c", &a[i][j]);
      p |= a[i][j] - '0';
      if (a[i][j] != '0') {
        for (k = 0; k < 3; k++) {
          A[i][j][0][k] +=
              A[i + yon[k][0]][j + yon[k][1]][0][k] + (a[i][j] == '2');
          A[i][j][1][k] +=
              A[i + yon[k][0]][j + yon[k][1]][1][k] + (a[i][j] == '3');
          A[i][j][2][k] +=
              A[i + yon[k][0]][j + yon[k][1]][2][k] + (a[i][j] == '1');
        }
      }
    }
  if (!p) {
    puts("0");
    return 0;
  }
  for (i = 1; i <= n; i++)
    for (j = n; j >= 1; j--) {
      if (a[i][j] != '0') {
        for (k = 3; k < 5; k++) {
          A[i][j][0][k] +=
              A[i + yon[k][0]][j + yon[k][1]][0][k] + (a[i][j] == '2');
          A[i][j][1][k] +=
              A[i + yon[k][0]][j + yon[k][1]][1][k] + (a[i][j] == '3');
          A[i][j][2][k] +=
              A[i + yon[k][0]][j + yon[k][1]][2][k] + (a[i][j] == '1');
        }
      }
    }
  for (i = n; i >= 1; i--)
    for (j = n; j >= 1; j--) {
      if (a[i][j] != '0') {
        for (k = 5; k < 6; k++) {
          A[i][j][0][k] +=
              A[i + yon[k][0]][j + yon[k][1]][0][k] + (a[i][j] == '2');
          A[i][j][1][k] +=
              A[i + yon[k][0]][j + yon[k][1]][1][k] + (a[i][j] == '3');
          A[i][j][2][k] +=
              A[i + yon[k][0]][j + yon[k][1]][2][k] + (a[i][j] == '1');
        }
      }
    }
  for (i = n; i >= 1; i--)
    for (j = 1; j <= n; j++) {
      if (a[i][j] != '0') {
        for (k = 6; k < 8; k++) {
          A[i][j][2][k] +=
              A[i + yon[k][0]][j + yon[k][1]][2][k] + (a[i][j] == '1');
          A[i][j][0][k] +=
              A[i + yon[k][0]][j + yon[k][1]][0][k] + (a[i][j] == '2');
          A[i][j][1][k] +=
              A[i + yon[k][0]][j + yon[k][1]][1][k] + (a[i][j] == '3');
        }
      }
    }
  pair<int, int> t;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) {
      if (a[i][j] == '0') continue;
      int mn = 1000000007;
      t = make_pair(0, 0);
      t.first -= (a[i][j] == '2') * 3;
      t.second -= (a[i][j] == '3') * 3;
      for (k = 0; k < 8; k += 2)
        mn = min(mn, A[i][j][0][k] + A[i][j][1][k] + A[i][j][2][k]);
      for (k = 0; k < 8; k += 2) {
        t.first +=
            A[i][j][0][k] - A[i + mn * yon[k][0]][j + mn * yon[k][1]][0][k];
        t.second +=
            A[i][j][1][k] - A[i + mn * yon[k][0]][j + mn * yon[k][1]][1][k];
      }
      v.push_back(t);
      t = make_pair(0, 0);
      t.first -= (a[i][j] == '2') * 3;
      t.second -= (a[i][j] == '3') * 3;
      mn = 1000000007;
      for (k = 1; k < 8; k += 2)
        mn = min(mn, A[i][j][0][k] + A[i][j][1][k] + A[i][j][2][k]);
      for (k = 1; k < 8; k += 2) {
        t.first +=
            A[i][j][0][k] - A[i + mn * yon[k][0]][j + mn * yon[k][1]][0][k];
        t.second +=
            A[i][j][1][k] - A[i + mn * yon[k][0]][j + mn * yon[k][1]][1][k];
      }
      v.push_back(t);
    }
  double l = 0;
  long long ans = 1;
  for (i = 1; i <= 2000; i++) {
    l += log2(3);
    for (j = 0; j <= min((int)l, 2000); j++) H[i][j] = 1;
  }
  sort(v.begin(), v.end(), cmp);
  t = v.back();
  for (i = 0; i < t.first; i++) ans = ans * 2 % 1000000007;
  for (i = 0; i < t.second; i++) ans = ans * 3 % 1000000007;
  cout << ans << '\n';
}
