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 dx[8] = {-1, 1, 0, 0, -1, -1, 1, 1},
          dy[8] = {0, 0, 1, -1, 1, -1, -1, 1};
const int N = 1e3 + 10;
const long long mod = 1e9 + 7;
double lg2, lg3;
char maze[N][N];
int lt[N][N], rt[N][N], up[N][N], dw[N][N], a[N][N];
int dp1[N][N], dp2[N][N], dp3[N][N], dp4[N][N];
double b[N][N];
double sc[N][N], sr[N][N];
int n, m;
void ac(int len, int x, int y) {
  sr[x][y - len]++;
  sr[x][y + len + 1]--;
  sc[x - len][y]++;
  sc[x + len + 1][y]--;
}
int main() {
  lg2 = log(2), lg3 = log(3);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%s", maze[i] + 1);
    for (int j = 1; j <= n; j++) {
      a[i][j] = maze[i][j] - '0';
      if (a[i][j] == 2) {
        b[i][j] = lg2;
      } else if (a[i][j] == 3) {
        b[i][j] = lg3;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (a[i][j]) {
        up[i][j] = up[i - 1][j] + 1;
        lt[i][j] = lt[i][j - 1] + 1;
        dp1[i][j] = dp1[i - 1][j - 1] + 1;
        dp2[i][j] = dp2[i - 1][j + 1] + 1;
      }
      sr[i][j] += sr[i][j - 1] + b[i][j], sc[i][j] += sc[i - 1][j] + b[i][j];
    }
  }
  for (int i = n; i >= 1; i--) {
    for (int j = n; j >= 1; j--) {
      if (a[i][j]) {
        dw[i][j] = dw[i + 1][j] + 1;
        rt[i][j] = rt[i][j + 1] + 1;
        dp3[i][j] = dp3[i + 1][j + 1] + 1;
        dp4[i][j] = dp4[i + 1][j - 1] + 1;
      }
    }
  }
  double mv = -1;
  int mx = 1, my = 1, op = 1, len = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (a[i][j]) {
        int cnt = min(min(lt[i][j - 1], rt[i][j + 1]),
                      min(up[i - 1][j], dw[i + 1][j]));
        double val = sr[i][j + cnt] - sr[i][j - cnt - 1] + sc[i + cnt][j] -
                     sc[i - cnt - 1][j] - b[i][j];
        if (val > mv) mv = val, mx = i, my = j, op = 1, len = cnt;
      }
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      sr[i][j] = sr[i - 1][j - 1] + b[i][j],
      sc[i][j] = sc[i - 1][j + 1] + b[i][j];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (a[i][j]) {
        int cnt1 = min(min(dp1[i - 1][j - 1], dp2[i - 1][j + 1]),
                       min(dp3[i + 1][j + 1], dp4[i + 1][j - 1]));
        double val = sr[i + cnt1][j + cnt1] - sr[i - cnt1 - 1][j - cnt1 - 1] +
                     sc[i + cnt1][j - cnt1] - sc[i - cnt1 - 1][j + cnt1 + 1] -
                     b[i][j];
        if (val > mv) mv = val, mx = i, my = j, op = 2, len = cnt1;
      }
    }
  }
  long long ans = 1;
  if (op == 1) {
    for (int i = 0; i < 4; i++) {
      int x = mx, y = my;
      for (int j = 0; j < len; j++) {
        x += dx[i], y += dy[i];
        ans = ans * a[x][y] % mod;
      }
    }
    ans = ans * a[mx][my] % mod;
  } else {
    for (int i = 4; i < 8; i++) {
      int x = mx, y = my;
      for (int j = 0; j < len; j++) {
        x += dx[i], y += dy[i];
        ans = ans * a[x][y] % mod;
      }
    }
    ans = ans * a[mx][my] % mod;
  }
  cout << ans << endl;
  return 0;
}
