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>
#pragma comment(linker, "/STACK:66777216")
using namespace std;
const int N = 1000 + 5;
const int T = 1000;
int n;
char c[N][N];
int a[N][N];
double b[N][N];
const int M = 8;
int dx[] = {1, -1, 0, 0, 1, -1, -1, 1};
int dy[] = {0, 0, 1, -1, 1, 1, -1, -1};
int si[] = {T, 1, 1, 1, T, 1, 1, T};
int fi[] = {1, T, T, T, 1, T, T, 1};
int sj[] = {1, 1, T, 1, T, T, 1, 1};
int fj[] = {T, T, 1, T, 1, 1, T, T};
int di[] = {-1, 1, 1, 1, -1, 1, 1, -1};
int dj[] = {1, 1, -1, 1, -1, -1, 1, 1};
int q[8][N][N];
double s[8][N][N];
const int MO = 1000000000 + 7;
int mul(int a, int b) { return ((long long)a * b) % MO; }
void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= (int)(n); ++i) scanf("%s", c[i]);
  for (int i = 1; i <= (int)(n); ++i)
    for (int j = 0; j < (int)(n); ++j) a[i][j + 1] = c[i][j] - '0';
  for (int i = 0; i < (int)(n + 2); ++i)
    for (int j = 0; j < (int)(n + 2); ++j)
      if (s[i][j] != 0)
        b[i][j] = (double)log((double)a[i][j]);
      else
        b[i][j] = -N * (double)log((double)3.0);
  for (int u = 0; u < (int)(8); ++u) {
    for (int i = si[u]; i != fi[u] + di[u]; i += di[u]) {
      for (int j = sj[u]; j != fj[u] + dj[u]; j += dj[u]) {
        if (a[i][j] == 0) {
          q[u][i][j] = 0;
          s[u][i][j] = 0.0;
        } else {
          q[u][i][j] = q[u][i + dx[u]][j + dy[u]] + 1;
          s[u][i][j] = s[u][i + dx[u]][j + dy[u]] + b[i][j];
        }
      }
    }
  }
  int ru = 0, rx = 1, ry = 1, rQ = 1;
  long double res = 0.0;
  for (int u = 0; u < 7; u += 4) {
    for (int i = 1; i <= (int)(n); ++i) {
      for (int j = 1; j <= (int)(n); ++j) {
        if (a[i][j] != 0) {
          int Q = q[u][i][j];
          for (int v = u + 1; v < u + 4; ++v) Q = min(Q, q[v][i][j]);
          long double ans = 0.0;
          for (int v = u; v < u + 4; ++v)
            ans = ans + s[v][i + dx[v]][j + dy[v]] -
                  s[v][i + Q * dx[v]][j + Q * dy[v]];
          ans += b[i][j];
          if (ans > res) {
            res = ans;
            rQ = Q;
            ru = u;
            rx = i;
            ry = j;
          }
        }
      }
    }
  }
  int ANS = 1;
  ANS = mul(ANS, a[rx][ry]);
  for (int v = ru; v < ru + 4; ++v) {
    int cx = rx + dx[v];
    int cy = ry + dy[v];
    for (int i = 1; i < rQ; ++i) {
      ANS = mul(ANS, a[cx][cy]);
      cx += dx[v];
      cy += dy[v];
    }
  }
  printf("%d\n", ANS);
}
void testgen() {
  FILE* f = fopen("input.txt", "w");
  fclose(f);
}
int main() {
  std::cout << std::fixed;
  std::cout.precision(8);
  std::cerr << std::fixed;
  std::cerr.precision(3);
  solve();
  return 0;
}
