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;
inline char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read() {
  int x = 0;
  char ch = getchar();
  bool positive = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') positive = 0;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return positive ? x : -x;
}
inline void write(int a) {
  if (a >= 10) write(a / 10);
  putchar('0' + a % 10);
}
inline void writeln(int a) {
  if (a < 0) {
    a = -a;
    putchar('-');
  }
  write(a);
  puts("");
}
const int mod = 1000000007;
const int N = 1005, xx[8] = {0, 0, 1, -1, 1, 1, -1, -1},
          yy[8] = {1, -1, 0, 0, 1, -1, 1, -1};
int n, dp[N][N][8], a[N][N], sum[N][N][8][2], tong[2];
pair<int, int> ans;
char ch[N];
inline void dfs(int x, int y, int id) {
  if (dp[x][y][id] != -1) return;
  if (a[x][y] == 0) {
    dp[x][y][id] = 0;
    return;
  }
  dfs(x - xx[id], y - yy[id], id);
  dp[x][y][id] = dp[x - xx[id]][y - yy[id]][id] + 1;
  for (int i = 0; i < 2; i++)
    sum[x][y][id][i] = sum[x - xx[id]][y - yy[id]][id][i];
  if (a[x][y] > 1) sum[x][y][id][a[x][y] - 2]++;
}
const double s1 = log(2), s2 = log(3);
inline void geng(pair<int, int> &x, pair<int, int> xx) {
  double t3 = s1 * x.first + s2 * x.second, t4 = s1 * xx.first + s2 * xx.second;
  if (t4 > t3) x = xx;
}
int main() {
  n = read();
  int sss = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%s", ch + 1);
    for (int j = 1; j <= n; j++) sss += (a[i][j] = ch[j] - '0');
  }
  if (!sss) {
    puts("0");
    return 0;
  }
  memset(dp, -1, sizeof(dp));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      for (int k = 0; k < 8; k++) dfs(i, j, k);
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++)
      if (a[i][j]) {
        int ttt = 1e9;
        for (int k = 0; k < 4; k++) {
          dfs(i, j, k);
          ttt = min(ttt, dp[i][j][k]);
        }
        tong[0] = tong[1] = 0;
        for (int k = 0; k < 4; k++) {
          for (int l = 0; l < 2; l++) {
            tong[l] += sum[i - xx[k]][j - yy[k]][k][l] -
                       sum[i - ttt * xx[k]][j - ttt * yy[k]][k][l];
          }
        }
        if (a[i][j] >= 2) tong[a[i][j] - 2]++;
        geng(ans, make_pair(tong[0], tong[1]));
        ttt = 1e9;
        for (int k = 4; k < 8; k++) {
          dfs(i, j, k);
          ttt = min(ttt, dp[i][j][k]);
        }
        tong[0] = tong[1] = 0;
        for (int k = 4; k < 8; k++) {
          for (int l = 0; l < 2; l++) {
            tong[l] += sum[i - xx[k]][j - yy[k]][k][l] -
                       sum[i - ttt * xx[k]][j - ttt * yy[k]][k][l];
          }
        }
        if (a[i][j] >= 2) tong[a[i][j] - 2]++;
        geng(ans, make_pair(tong[0], tong[1]));
      }
  }
  int an = 1;
  for (int i = 0; i < ans.first; i++) an = an * 2 % mod;
  for (int i = 0; i < ans.second; i++) an = (long long)an * 3 % mod;
  cout << an << endl;
}
