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;
struct node {
  int l, e, s;
  node() {}
  node(int a, int b, int c) : l(a), e(b), s(c) {}
} f[1010][1010][8];
char s[1010];
int g[8][2] = {0, -1, -1, -1, -1, 0, -1, 1, 0, 1, 1, 1, 1, 0, 1, -1};
int a[1010][1010];
node get(int x, node k) {
  node ret = node(0, 0, 0);
  if (x == 0) {
    return ret;
  }
  ret = k;
  ret.l++;
  if (x == 2) {
    ret.e++;
  } else if (x == 3) {
    ret.s++;
  }
  return ret;
}
bool pd(int a, int b, int c, int d) {
  return (a - c) * log(2) + (b - d) * log(3) >= 0;
}
int main() {
  int i, j, k, h, n, x = 0, y = 0, xx, yy, mx = 0;
  long long ans;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%s", s);
    for (j = 1; j <= n; j++) {
      a[i][j] = s[j - 1] - '0';
      mx = max(mx, a[i][j]);
    }
  }
  memset(f, 0, sizeof(0));
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (a[i][j]) {
        for (k = 0; k < 4; k++) {
          f[i][j][k] =
              get(a[i + g[k][0]][j + g[k][1]], f[i + g[k][0]][j + g[k][1]][k]);
        }
      }
    }
  }
  for (i = n; i >= 1; i--) {
    for (j = n; j >= 1; j--) {
      if (a[i][j]) {
        for (k = 4; k < 8; k++) {
          f[i][j][k] =
              get(a[i + g[k][0]][j + g[k][1]], f[i + g[k][0]][j + g[k][1]][k]);
        }
      }
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (a[i][j]) {
        xx = yy = 0;
        if (a[i][j] == 2) {
          xx++;
        } else if (a[i][j] == 3) {
          yy++;
        }
        k = min(f[i][j][0].l,
                min(f[i][j][2].l, min(f[i][j][4].l, f[i][j][6].l)));
        for (h = 0; h <= 6; h += 2) {
          xx += f[i][j][h].e - f[i + k * g[h][0]][j + k * g[h][1]][h].e;
          yy += f[i][j][h].s - f[i + k * g[h][0]][j + k * g[h][1]][h].s;
        }
        if (pd(xx, yy, x, y)) {
          x = xx, y = yy;
        }
        xx = yy = 0;
        if (a[i][j] == 2) {
          xx++;
        } else if (a[i][j] == 3) {
          yy++;
        }
        k = min(f[i][j][1].l,
                min(f[i][j][3].l, min(f[i][j][5].l, f[i][j][7].l)));
        for (h = 1; h <= 7; h += 2) {
          xx += f[i][j][h].e - f[i + k * g[h][0]][j + k * g[h][1]][h].e;
          yy += f[i][j][h].s - f[i + k * g[h][0]][j + k * g[h][1]][h].s;
        }
        if (pd(xx, yy, x, y)) {
          x = xx, y = yy;
        }
      }
    }
  }
  if (x == 0 && y == 0) {
    printf("%d\n", mx);
  } else {
    long long ans = 1;
    for (i = 1; i <= x; i++) {
      ans *= 2;
      ans %= mod;
    }
    for (j = 1; j <= y; j++) {
      ans *= 3;
      ans %= mod;
    }
    printf("%I64d\n", ans);
  }
}
