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;
long long mod = 1000000007LL;
int main() {
  vector<vector<int> > left, right, up, down;
  vector<vector<int> > grid;
  vector<vector<long double> > l, r, u, d;
  int n;
  cin >> n;
  grid.assign(n + 10, vector<int>(n + 10, 0));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      char c;
      cin >> c;
      grid[i][j] = (int)(c - '0');
    }
  }
  left.assign(n + 10, vector<int>(n + 10, 0));
  right = up = down = left;
  l.assign(n + 10, vector<long double>(n + 10, (double)0.0));
  r = u = d = l;
  vector<long double> lg(10, 0.0);
  for (int i = 1; i < 10; i++) {
    lg[i] = (long double)log((long double)i);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (grid[i][j] == 0) {
      } else {
        left[i][j] = left[i][j - 1] + 1;
      }
      l[i][j] = l[i][j - 1] + lg[grid[i][j]];
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = n; j >= 1; j--) {
      if (grid[i][j] != 0) {
        right[i][j] = right[i][j + 1] + 1;
      }
      r[i][j] = r[i][j + 1] + lg[grid[i][j]];
    }
  }
  for (int j = 1; j <= n; j++) {
    for (int i = 1; i <= n; i++) {
      if (grid[i][j] != 0) {
        up[i][j] = up[i - 1][j] + 1;
      }
      u[i][j] = u[i - 1][j] + lg[grid[i][j]];
    }
  }
  for (int j = 1; j <= n; j++) {
    for (int i = n; i >= 1; i--) {
      if (grid[i][j] != 0) {
        down[i][j] = down[i + 1][j] + 1;
      }
      d[i][j] = d[i + 1][j] + lg[grid[i][j]];
    }
  }
  long double ans = -1.0;
  int pr = 1, pc = 1, pd = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (grid[i][j] == 0) continue;
      int d = min(min(up[i][j], down[i][j]), min(left[i][j], right[i][j]));
      long double temp = u[i + d - 1][j] - u[i - d][j] + l[i][j + d - 1] -
                         l[i][j - d] - lg[grid[i][j]];
      if (ans - temp < 0) {
        ans = temp;
        pr = i;
        pc = j;
        pd = d;
      }
    }
  }
  long long a = 1LL;
  for (int i = 1; i < pd; i++) {
    a = (a * (long long)grid[pr][pc - i]) % mod;
    a = (a * (long long)grid[pr][pc + i]) % mod;
    a = (a * (long long)grid[pr + i][pc]) % mod;
    a = (a * (long long)grid[pr - i][pc]) % mod;
  }
  a = (a * (long long)grid[pr][pc]) % mod;
  up.assign(n + 10, vector<int>(n + 10, 0));
  down = left = right = up;
  l.assign(n + 10, vector<long double>(n + 10, 0.0));
  r = u = d = l;
  for (int lvl = 2; lvl <= 2 * n; lvl++) {
    for (int i = max(1, lvl - n); i <= min(n, lvl - 1); i++) {
      int j = lvl - i;
      if (grid[i][j] != 0) {
        up[i][j] = up[i - 1][j - 1] + 1;
        right[i][j] = right[i - 1][j + 1] + 1;
      }
      r[i][j] = r[i - 1][j + 1] + lg[grid[i][j]];
      u[i][j] = u[i - 1][j - 1] + lg[grid[i][j]];
    }
  }
  for (int lvl = 2 * n; lvl >= 2; lvl--) {
    for (int i = min(n, lvl - 1); i >= max(1, lvl - n); i--) {
      int j = lvl - i;
      if (grid[i][j] != 0) {
        left[i][j] = left[i + 1][j - 1] + 1;
        down[i][j] = down[i + 1][j + 1] + 1;
      }
      l[i][j] = l[i + 1][j - 1] + lg[grid[i][j]];
      d[i][j] = d[i + 1][j + 1] + lg[grid[i][j]];
    }
  }
  double ans2 = -1.0;
  for (int lvl = 2; lvl <= 2 * n; lvl++) {
    for (int i = max(1, lvl - n); i <= min(n, lvl - 1); i++) {
      int j = lvl - i;
      int d = min(min(up[i][j], down[i][j]), min(left[i][j], right[i][j]));
      long double temp = u[i + d - 1][j + d - 1] - u[i - d][j - d] +
                         r[i + d - 1][j - d + 1] - r[i - d][j + d] -
                         lg[grid[i][j]];
      if (ans2 - temp < 0) {
        ans2 = temp;
        pr = lvl;
        pc = i;
        pd = d;
      }
    }
  }
  long long b = 1LL;
  for (int i = 1; i < pd; i++) {
    b = (b * (long long)grid[pc + i][pr - pc + i]) % mod;
    b = (b * (long long)grid[pc + i][pr - pc - i]) % mod;
    b = (b * (long long)grid[pc - i][pr - pc - i]) % mod;
    b = (b * (long long)grid[pc - i][pr - pc + i]) % mod;
  }
  b = (b * (long long)grid[pc][pr - pc]) % mod;
  if (ans2 > ans) a = b;
  cout << a << endl;
  return 0;
}
