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 mod = 1e9 + 7;
const int N = 1005;
const long double two = log2(2), three = log2(3);
int n;
char s[N][N];
bool all_zero() {
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (s[i][j] != '0') return false;
    }
  }
  return true;
}
int L[N][N], R[N][N], U[N][N], D[N][N];
long double rs[N][N], cs[N][N];
int r2[N][N], r3[N][N], c2[N][N], c3[N][N];
inline long double weight(char c) {
  if (c == '2') return two;
  if (c == '3') return three;
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%s", s[i] + 1);
  }
  if (all_zero()) {
    puts("0");
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      L[i][j] = s[i][j] == '0' ? j : L[i][j - 1];
      rs[i][j] = rs[i][j - 1] + weight(s[i][j]);
      r2[i][j] = r2[i][j - 1] + (s[i][j] == '2');
      r3[i][j] = r3[i][j - 1] + (s[i][j] == '3');
    }
    for (int j = n; j >= 1; --j) {
      R[i][j] = s[i][j] == '0' ? j : R[i][j + 1];
    }
  }
  for (int j = 1; j <= n; ++j) {
    for (int i = 1; i <= n; ++i) {
      U[i][j] = s[i][j] == '0' ? i : U[i - 1][j];
      cs[i][j] = cs[i - 1][j] + weight(s[i][j]);
      c2[i][j] = c2[i - 1][j] + (s[i][j] == '2');
      c3[i][j] = c3[i - 1][j] + (s[i][j] == '3');
    }
    for (int i = n; i >= 1; --i) {
      D[i][j] = s[i][j] == '0' ? i : D[i + 1][j];
    }
  }
  int A = 0, B = 0;
  long double value = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (s[i][j] == '0') continue;
      int d = min(i - 1, n - i);
      d = min(d, min(j - 1, n - j));
      if (U[i][j]) d = min(d, i - U[i][j] - 1);
      if (D[i][j]) d = min(d, D[i][j] - i - 1);
      if (L[i][j]) d = min(d, j - L[i][j] - 1);
      if (R[i][j]) d = min(d, R[i][j] - j - 1);
      long double v =
          rs[i][j + d] - rs[i][j - d - 1] + cs[i + d][j] - cs[i - d - 1][j];
      v -= weight(s[i][j]);
      if (v > value) {
        value = v;
        A = r2[i][j + d] - r2[i][j - d - 1] + c2[i + d][j] - c2[i - d - 1][j];
        if (s[i][j] == '2') --A;
        B = r3[i][j + d] - r3[i][j - d - 1] + c3[i + d][j] - c3[i - d - 1][j];
        if (s[i][j] == '3') --B;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      L[i][j] = s[i][j] == '0' ? 0 : L[i - 1][j - 1] + 1;
      rs[i][j] = rs[i - 1][j - 1] + weight(s[i][j]);
      r2[i][j] = r2[i - 1][j - 1] + (s[i][j] == '2');
      r3[i][j] = r3[i - 1][j - 1] + (s[i][j] == '3');
    }
  }
  for (int i = n; i >= 1; --i) {
    for (int j = n; j >= 1; --j) {
      R[i][j] = s[i][j] == '0' ? 0 : R[i + 1][j + 1] + 1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      U[i][j] = s[i][j] == '0' ? 0 : U[i - 1][j + 1] + 1;
      cs[i][j] = cs[i - 1][j + 1] + weight(s[i][j]);
      c2[i][j] = c2[i - 1][j + 1] + (s[i][j] == '2');
      c3[i][j] = c3[i - 1][j + 1] + (s[i][j] == '3');
    }
  }
  for (int i = n; i >= 1; --i) {
    for (int j = 1; j <= n; ++j) {
      D[i][j] = s[i][j] == '0' ? 0 : D[i + 1][j - 1] + 1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (s[i][j] == '0') continue;
      int d = min(i - 1, n - i);
      d = min(d, min(j - 1, n - j));
      d = min(d, L[i][j] - 1);
      d = min(d, R[i][j] - 1);
      d = min(d, U[i][j] - 1);
      d = min(d, D[i][j] - 1);
      long double v = rs[i + d][j + d] - rs[i - d - 1][j - d - 1] +
                      cs[i + d][j - d] - cs[i - d - 1][j + d + 1];
      v -= weight(s[i][j]);
      if (v > value) {
        value = v;
        A = r2[i + d][j + d] - r2[i - d - 1][j - d - 1] + c2[i + d][j - d] -
            c2[i - d - 1][j + d + 1];
        if (s[i][j] == '2') --A;
        B = r3[i + d][j + d] - r3[i - d - 1][j - d - 1] + c3[i + d][j - d] -
            c3[i - d - 1][j + d + 1];
        if (s[i][j] == '3') --B;
      }
    }
  }
  int ans = 1;
  for (int i = 0; i < A; ++i) ans = ans * 2 % mod;
  for (int i = 0; i < B; ++i) ans = ans * 3ll % mod;
  printf("%d\n", ans);
  return 0;
}
