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 N = 1003;
const int D = 8;
const int MOD = 1000000007;
int dx[] = {1, 1, 1, 0, 0, -1, -1, -1};
int dy[] = {1, 0, -1, 1, -1, 1, 0, -1};
char grid[N][N];
int first_zero[D][N][N];
double sum[D][N][N];
int n;
inline bool IsValid(int i, int j) { return min(i, j) >= 0 && max(i, j) < n; }
inline bool IsRotated(int dx, int dy) { return min(abs(dx), abs(dy)) != 0; }
void GetMaxVal(int i, int j, int rot, double& lg, int& k) {
  k = n;
  for (int d = 0; d < D; ++d) {
    if (rot != IsRotated(dx[d], dy[d])) continue;
    k = min(k, first_zero[d][i][j]);
  }
  lg = log2(grid[i][j]);
  for (int d = 0; d < D; ++d) {
    if (rot != IsRotated(dx[d], dy[d])) continue;
    lg += sum[d][i + k * dx[d]][j + k * dy[d]] - sum[d][i][j];
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      cin >> grid[i][j];
      grid[i][j] -= '0';
    }
  }
  for (int d = 0; d < 8; ++d) {
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        if (IsValid(i - dx[d], j - dy[d])) continue;
        int last_zero = -1;
        int x = i, y = j;
        double last_sum = 0;
        for (int k = 0; IsValid(x, y); ++k, x += dx[d], y += dy[d]) {
          char curr = grid[x][y];
          double lg = 0;
          if (curr == 0) {
            last_zero = k;
          } else {
            lg = log2(curr);
          }
          first_zero[d][x][y] = k - last_zero - 1;
          sum[d][x][y] = last_sum + lg;
          last_sum = sum[d][x][y];
        }
      }
    }
  }
  double max_log = 0;
  int x = 0, y = 0;
  bool rotated = false;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j)
      for (int rot = 0; rot < 2; ++rot) {
        if (grid[i][j] == 0) continue;
        double lg;
        int k;
        GetMaxVal(i, j, rot, lg, k);
        if (lg > max_log) {
          max_log = lg;
          x = i;
          y = j;
          rotated = rot;
        }
      }
  }
  double lg;
  int k;
  GetMaxVal(x, y, rotated, lg, k);
  int ans = grid[x][y];
  for (int d = 0; d < D; ++d) {
    if (rotated != IsRotated(dx[d], dy[d])) continue;
    for (int i = 1; i <= k; ++i) {
      ans = 1LL * ans * grid[x + i * dx[d]][y + i * dy[d]] % MOD;
    }
  }
  cout << ans;
}
